IB Applications and Interpretation SL · Differential and integral calculus · End of topic

IB exam questions for topic 7

86 exercises with solutions · By Álvaro López

Past IB exam questions grouped by type and, within each type, from easiest to hardest. Try each one before opening the solution.

Level

Paper

Solutions

Derivatives and differentiation rules

Exercise 1 BasicIB · P1 TZ0 May 2006

The function $f (x)$ is defined by $f (x) = 3 x^{4} + \frac{2}{x} - \frac{x}{4} + 1$ , $(x \neq = 0)$ .

a) Calculate the second derivative $f^{''} (x)$ .

b) Find the value of $f^{''} (x)$ at the point $(1, \frac{23}{4})$ .

Show answer

Answer: a) $f^{''} (x) = 36 x^{2} + \frac{4}{x ^{3}}$ ; b) $f^{''} (1) = 40$

Show working and answer

We write $f (x) = 3 x^{4} + 2 x^{- 1} - \frac{1}{4} x + 1$ .
First derivative: $f^{'} (x) = 12 x^{3} - \frac{2}{x ^{2}} - \frac{1}{4}$ .
Second derivative: $f^{''} (x) = 36 x^{2} + \frac{4}{x ^{3}} .$

We substitute $x = 1$ : $f^{''} (1) = 36 + 4 = 40.$

Answer: a) $f^{''} (x) = 36 x^{2} + \frac{4}{x ^{3}}$ ; b) $f^{''} (1) = 40$

Exercise 2 BasicIB · P1 TZ0 November 2009

Let $f (x) = 2 x^{2} + x - 6$ .

a) Find $f^{'} (x)$ .

b) Find the value of $f^{'} (- 3)$ .

c) Find the value of $x$ for which $f^{'} (x) = 0$ .

Show answer

Answer: a) $f^{'} (x) = 4 x + 1$ ; b) $- 11$ ; c) $x = - \frac{1}{4} = - 0.25$

Show working and answer

a) $f^{'} (x) = 4 x + 1.$

b) $f^{'} (- 3) = 4 (- 3) + 1 = - 11.$

$4 x + 1 = 0 \Rightarrow x = - \frac{1}{4}$ .

Answer: a) $f^{'} (x) = 4 x + 1$ ; b) $- 11$ ; c) $x = - \frac{1}{4} = - 0.25$

Exercise 3 BasicIB · P2 TZ0 November 2013

Let $f (x) = 3 x^{4} - \frac{1}{2}$ .

a) Find $f^{'} (x)$ .

b) Find $\int f (x) d x$ .

Show answer

Answer: a) $f^{'} (x) = \frac{4}{3} x^{1/3}$ ; b) $\int f (x) d x = \frac{3}{7} x^{7/3} - \frac{1}{2} x + C$

Show working and answer

Rewrite using a fractional index: $3 x^{4} = x^{4/3}$ , so $f (x) = x^{4/3} - \frac{1}{2}$ .
Differentiate: $f^{'} (x) = \frac{4}{3} x^{4/3 - 1} = \frac{4}{3} x^{1/3} .$

Integrate each term: $\int x^{4/3} d x = \frac{x ^{7/3}}{7/3} = \frac{3}{7} x^{7/3}$ and $\int \frac{1}{2} d x = \frac{1}{2} x$ .
Hence $\int f (x) d x = \frac{3}{7} x^{7/3} - \frac{1}{2} x + C .$

Answer: a) $f^{'} (x) = \frac{4}{3} x^{1/3}$ ; b) $\int f (x) d x = \frac{3}{7} x^{7/3} - \frac{1}{2} x + C$

Exercise 4 MediumIB · P1 TZ0 November 2019

The following figure shows the curve $y = \frac{x ^{2}}{2} + \frac{2 a}{x}$ , $x \neq = 0$ .

Curve y = x²/2 + 2a/x (drawn with a = 1) with two branches separated by the vertical axis; the left branch cuts the horizontal axis to the left of the origin and the right branch has a local minimum to the right of the vertical axis

The equation of the vertical asymptote of this curve is $x = k$ .

a) Write down the value of $k$ .

b) Find $\frac{d y}{d x}$ .

At the point where $x = 2$ , the gradient of the tangent to the curve is equal to $0.5$ .

c) Find the value of $a$ .

Show answer

Answer: a) $k = 0$ ; b) $\frac{d y}{d x} = x - \frac{2 a}{x ^{2}}$ ; c) $a = 3$

Show working and answer

The curve is not defined at $x = 0$ , and the term $\frac{2 a}{x}$ becomes infinite as $x$ approaches $0$ .
The vertical asymptote is the $y$ -axis: $k = 0$ .

We write $y = \frac{1}{2} x^{2} + 2 a x^{- 1}$ and differentiate term by term: $\frac{d y}{d x} = x - \frac{2 a}{x ^{2}} .$

We substitute $x = 2$ into the derivative: $\frac{d y}{d x} = 2 - \frac{2 a}{4} = 2 - \frac{a}{2}$ .
We impose that the gradient is $0.5$ : $2 - \frac{a}{2} = 0.5 \Rightarrow \frac{a}{2} = 1.5 \Rightarrow a = 3.$

Answer: a) $k = 0$ ; b) $\frac{d y}{d x} = x - \frac{2 a}{x ^{2}}$ ; c) $a = 3$

Exercise 5 MediumIB · P1 TZ2 May 2018

Consider the function $f (x) = \frac{x ^{4}}{4}$ .

a) Find $f^{'} (x)$ .

b) Find the gradient of the graph of $f$ at $x = - \frac{1}{2}$ .

c) Find the $x$ -coordinate of the point at which the normal to the graph of $f$ has gradient $- \frac{1}{8}$ .

Show answer

Answer: a) $f^{'} (x) = x^{3}$ ; b) $- \frac{1}{8}$ ; c) $x = 2$

Show working and answer

$f^{'} (x) = \frac{4 x ^{3}}{4} = x^{3}$ .

$f^{'} (- \frac{1}{2}) = (- \frac{1}{2})^{3} = - \frac{1}{8}$ .

The normal has gradient $- \frac{1}{8}$ , so the tangent at that point has gradient $8$ (perpendicular gradients: $m_{1} m_{2} = - 1$ ).
We equate the derivative to $8$ : $x^{3} = 8 \Rightarrow x = 2.$

Answer: a) $f^{'} (x) = x^{3}$ ; b) $- \frac{1}{8}$ ; c) $x = 2$

Exercise 6 MediumIB · P2 TZ0 November 2007

a) Sketch the graph of the function $f (x) = x^{3} - 2 x^{2} + x - 3$ for values of $x$ between $- 2$ and $4$ , giving the intercepts with both axes.

b) On the same diagram, sketch the line $y = 7 - 2 x$ and find the coordinates of the point of intersection of the line with the curve.

c) Find the value of the gradient of the curve at $x = 1.7$ .

Show answer

Answer: a) graph with intercepts $(0, - 3)$ and $(2.17, 0)$ ; b) $(2.45, 2.11)$ ; c) gradient $2.87$

Show working and answer

Using the GDC, draw $f (x) = x^{3} - 2 x^{2} + x - 3$ for $- 2 \leq x \leq 4$ .
$y$ -intercept: $f (0) = - 3$ , that is, $(0, - 3)$ .
$x$ -intercept: $f (x) = 0$ gives $x \approx 2.17$ , that is, $(2.17, 0)$ .

Graph of f(x) = x³ − 2x² + x − 3 for −2 ≤ x ≤ 4, and of the line y = 7 − 2x, which intersect at a single point to the right of the y-axis; the curve passes through (0, −3) and crosses the x-axis at x ≈ 2.17

Draw the line $y = 7 - 2 x$ on the same axes (it passes through $(0, 7)$ and $(3.5, 0)$ ).
Using the GDC, the point of intersection has $x \approx 2.45$ . Then $y = 7 - 2 (2.45) \approx 2.11$ .
The line and the curve intersect at $(2.45, 2.11)$ .

Differentiate: $f^{'} (x) = 3 x^{2} - 4 x + 1$ .
Substitute $x = 1.7$ : $f^{'} (1.7) = 3 (1.7)^{2} - 4 (1.7) + 1 = 8.67 - 6.8 + 1 = 2.87.$

Answer: a) graph with intercepts $(0, - 3)$ and $(2.17, 0)$ ; b) $(2.45, 2.11)$ ; c) gradient $2.87$

Exercise 7 MediumIB · P1 TZ1 May 2016

A company sells fruit juices in cylindrical cans, each of which has a volume of $340 cm^{3}$ . The surface area of a can is $A cm^{2}$ and is given by the formula $A = 2 π r^{2} + \frac{680}{r}$ , where $r$ is the radius of the can, in cm.

To reduce the cost of a can, its surface area must be minimised.

a) Calculate $\frac{d A}{d r}$ .

b) Calculate the value of $r$ that minimises the surface area of a can.

Show answer

Answer: a) $\frac{d A}{d r} = 4 π r - \frac{680}{r ^{2}}$ ; b) $r = 3 \frac{170}{π} \approx 3.78$ cm

Show working and answer

Write $A = 2 π r^{2} + 680 r^{- 1}$ and differentiate: $\frac{d A}{d r} = 4 π r - \frac{680}{r ^{2}} .$

The minimum occurs where the derivative is zero: $4 π r - \frac{680}{r ^{2}} = 0 \Rightarrow 4 π r^{3} = 680 \Rightarrow r^{3} = \frac{170}{π} .$
Therefore $r = 3 \frac{170}{π} \approx 3.78 cm .$
It is a minimum because $\frac{d A}{d r} < 0$ for smaller $r$ and $\frac{d A}{d r} > 0$ for larger $r$ (or because $A^{''} = 4 π + \frac{1360}{r ^{3}} > 0$ ).

Answer: a) $\frac{d A}{d r} = 4 π r - \frac{680}{r ^{2}}$ ; b) $r = 3 \frac{170}{π} \approx 3.78$ cm

Exercise 8 MediumIB · P1 TZ2 May 2016

Consider the function $f (x) = x^{3} - 3 x^{2} + 2 x + 2$ . Part of the graph of $f$ is shown below.

Graph of f(x) = x³ − 3x² + 2x + 2 with a reference grid; the axes run from −2 to 4 in x and from −15 to 25 in y

a) Find $f^{'} (x)$ .

b) There are two points at which the gradient of the graph of $f$ is equal to 11. Find the $x$ -coordinate of these points.

Show answer

Answer: a) $f^{'} (x) = 3 x^{2} - 6 x + 2$ ; b) $x = - 1$ and $x = 3$

Show working and answer

Differentiate each term of $f (x) = x^{3} - 3 x^{2} + 2 x + 2$ : $f^{'} (x) = 3 x^{2} - 6 x + 2.$

The gradient of the graph at a point is $f^{'} (x)$ , so we set $f^{'} (x) = 11$ : $3 x^{2} - 6 x + 2 = 11 \Rightarrow 3 x^{2} - 6 x - 9 = 0 \Rightarrow x^{2} - 2 x - 3 = 0.$
Factorise: $(x - 3) (x + 1) = 0.$
The $x$ -coordinates of these two points are $x = - 1$ and $x = 3$ .

Answer: a) $f^{'} (x) = 3 x^{2} - 6 x + 2$ ; b) $x = - 1$ and $x = 3$

Exercise 9 MediumIB · P1 TZ0 November 2016

The equation of a curve is $y = \frac{1}{2} x^{4} - \frac{3}{2} x^{2} + 7$ .

a) Find $\frac{d y}{d x}$ .

The gradient of the tangent to the curve at a point P is $- 10$ .

b) Find the coordinates of P.

Show answer

Answer: a) $\frac{d y}{d x} = 2 x^{3} - 3 x$ ; b) $P (- 2, 9)$

Show working and answer

Differentiate $y = \frac{1}{2} x^{4} - \frac{3}{2} x^{2} + 7$ : $\frac{d y}{d x} = 2 x^{3} - 3 x .$

Set the gradient equal to $- 10$ : $2 x^{3} - 3 x = - 10 \Rightarrow 2 x^{3} - 3 x + 10 = 0.$
Using the calculator (or by trying values) we see that $x = - 2$ is a solution: $2 (- 8) + 6 + 10 = 0$ .
Factorising, $2 x^{3} - 3 x + 10 = (x + 2) (2 x^{2} - 4 x + 5)$ , and the second factor has no real roots (its discriminant is $16 - 40 < 0$ ). So $x = - 2$ is the only solution.
We calculate the $y$ -coordinate: $y = \frac{1}{2} (- 2)^{4} - \frac{3}{2} (- 2)^{2} + 7 = 8 - 6 + 7 = 9.$
The point is $P (- 2, 9)$ .

Answer: a) $\frac{d y}{d x} = 2 x^{3} - 3 x$ ; b) $P (- 2, 9)$

Exercise 10 MediumIB · P1 TZ2 May 2013

A curve is described by the function $f (x) = 3 x - \frac{2}{x ^{2}}$ , $x \neq = 0$ .

a) Find $f^{'} (x)$ .

The gradient of the curve at the point $A$ is 35.

b) Find the $x$ -coordinate of the point $A$ .

Show answer

Answer: a) $f^{'} (x) = 3 + \frac{4}{x ^{3}}$ ; b) $x = \frac{1}{2} = 0.5$

Show working and answer

Write $f (x) = 3 x - 2 x^{- 2}$ and differentiate: $f^{'} (x) = 3 + 4 x^{- 3} = 3 + \frac{4}{x ^{3}} .$

Set the gradient equal to $35$ : $3 + \frac{4}{x ^{3}} = 35 \Rightarrow \frac{4}{x ^{3}} = 32 \Rightarrow x^{3} = \frac{1}{8} .$
Hence $x = \frac{1}{2} = 0.5$ .

Answer: a) $f^{'} (x) = 3 + \frac{4}{x ^{3}}$ ; b) $x = \frac{1}{2} = 0.5$

Exercise 11 MediumIB · P1 TZ0 November 2014

Consider the curve $y = x^{2} + \frac{a}{x} - 1$ , $x \neq = 0$ .

a) Calculate $\frac{d y}{d x}$ .

The gradient of the tangent to the curve is $- 14$ when $x = 1$ .

b) Find the value of $a$ .

Show answer

Answer: a) $\frac{d y}{d x} = 2 x - \frac{a}{x ^{2}}$ ; b) $a = 16$

Show working and answer

Write $y = x^{2} + a x^{- 1} - 1$ and differentiate: $\frac{d y}{d x} = 2 x - \frac{a}{x ^{2}} .$

The gradient at $x = 1$ is $- 14$ : $2 \cdot 1 - \frac{a}{1} = - 14 \Rightarrow 2 - a = - 14.$
Hence $a = 16$ .

Answer: a) $\frac{d y}{d x} = 2 x - \frac{a}{x ^{2}}$ ; b) $a = 16$

Tangent and normal lines

Exercise 12 MediumIB · P2 TZ0 November 2018

Consider the function $f (x) = \frac{27}{x ^{2}} - 16 x$ , $x \neq = 0$ .

a) Sketch the approximate graph of $y = f (x)$ , for $- 4 \leq x \leq 3$ and $- 50 \leq y \leq 100$ .

b) Use your graphic display calculator to find

(i) the zero of $f (x)$ ;

(ii) the coordinates of the local minimum point;

(iii) the equation of the tangent to the graph of $y = f (x)$ at the point $(- 2; 38.75)$ .

Give your answer in the form $y = m x + c$ .

c) On the same axes, sketch the graph of the function $g (x) = 10 x + 40$ .

d) Solve the equation $f (x) = g (x)$ .

Show answer

Answer: a) graph with vertical asymptote $x = 0$ and local minimum at $x = - 1.5$ ; b) (i) $x \approx 1.19$ ; (ii) $(- 1.5, 36)$ ; (iii) $y = - 9.25 x + 20.25$ ; c) line through $(- 4, 0)$ and $(3, 70)$ ; d) $x \approx 0.684$

Show working and answer

Using the GDC, draw $f (x) = \frac{27}{x ^{2}} - 16 x$ in the window $- 4 \leq x \leq 3$ , $- 50 \leq y \leq 100$ . It has a vertical asymptote at $x = 0$ (the curve rises without bound on both sides), a local minimum for $x < 0$ and, for $x > 0$ , it falls to a value close to $- 45$ at $x = 3$ .

Graphs of f(x) = 27/x² − 16x (two branches on either side of the vertical axis) and g(x) = 10x + 40 on a grid, for −4 ≤ x ≤ 3 and −50 ≤ y ≤ 100

(i) Using the GDC, the zero of $f$ is $x \approx 1.19$ . (Algebraically, $27 = 16 x^{3}$ and $x = 3 27/16$ .)
(ii) Using the GDC, the local minimum is $(- 1.5, 36)$ .
(iii) Using the GDC, the tangent at $(- 2; 38.75)$ has gradient $f^{'} (- 2) = - 9.25$ .
Equation of the tangent: $y - 38.75 = - 9.25 (x + 2) \Rightarrow y = - 9.25 x + 20.25.$

The line $g (x) = 10 x + 40$ passes through $(- 4, 0)$ and $(3, 70)$ ; draw it on the same axes.

Set $f (x) = g (x)$ : $\frac{27}{x ^{2}} - 16 x = 10 x + 40 \Rightarrow 26 x^{3} + 40 x^{2} - 27 = 0.$
Using the GDC, the only solution is $x \approx 0.684.$

Exercise 13 MediumIB · P1 TZ1 May 2009

The straight line, $L$ , has the equation $2 y - 27 x - 9 = 0$ .

a) Find the gradient of $L$ .

Sarah wants to draw the tangent to $f (x) = x^{4}$ that is parallel to $L$ .

b) Write down $f^{'} (x)$ .

c) (i) Find the $x$ -coordinate of the point at which the tangent should be drawn.

(ii) Write down the value of $f (x)$ at this point.

Show answer

Answer: a) $\frac{27}{2} = 13.5$ ; b) $f^{'} (x) = 4 x^{3}$ ; c) (i) $x = 1.5$ ; (ii) $f (1.5) = \frac{81}{16} \approx 5.06$

Show working and answer

Solve $2 y - 27 x - 9 = 0$ for $y$ : $y = \frac{27}{2} x + \frac{9}{2} .$
The gradient of $L$ is $\frac{27}{2} = 13.5$ .

Differentiate $f (x) = x^{4}$ : $f^{'} (x) = 4 x^{3} .$

(i) The tangent is parallel to $L$ if it has the same gradient: $4 x^{3} = \frac{27}{2} \Rightarrow x^{3} = \frac{27}{8} \Rightarrow x = \frac{3}{2} = 1.5.$
(ii) We calculate the value of $f$ at that point: $f (\frac{3}{2}) = (\frac{3}{2})^{4} = \frac{81}{16} \approx 5.06.$

Answer: a) $\frac{27}{2} = 13.5$ ; b) $f^{'} (x) = 4 x^{3}$ ; c) (i) $x = 1.5$ ; (ii) $f (1.5) = \frac{81}{16} \approx 5.06$

Exercise 14 MediumIB · P2 TZ1 May 2016

A function, $f$ , is given by $f (x) = 4 \times 2^{- x} + 1.5 x - 5$ .

a) Calculate $f (0)$ .

b) Use your graphic display calculator (GDC) to solve $f (x) = 0$ .

c) Draw the graph of $y = f (x)$ for $- 2 \leq x \leq 6$ and $- 4 \leq y \leq 10$ , showing the intercepts with the $x$ -axis and the $y$ -axis. Use a scale of 2 cm to represent 2 units on both the horizontal $x$ -axis and the vertical $y$ -axis.

The function $f$ is the derivative of a function $g$ . It is known that $g (1) = 3$ .

d) (i) Calculate $g^{'} (1)$ .

(ii) Find the equation of the tangent to the graph of $y = g (x)$ at $x = 1$ . Give your answer in the form $y = m x + c$ .

Show answer

Answer: a) $f (0) = - 1$ ; b) $x \approx - 0.538$ and $x = 3$ ; c) graph with intercepts $(- 0.538; 0)$ , $(3; 0)$ and $(0; - 1)$ ; d) (i) $g^{'} (1) = - 1.5$ ; (ii) $y = - 1.5 x + 4.5$

Show working and answer

Substitute $x = 0$ : $f (0) = 4 \cdot 2^{0} + 0 - 5 = 4 - 5 = - 1.$

On the GDC, we solve $4 \cdot 2^{- x} + 1.5 x - 5 = 0$ and obtain two solutions.
They are $x \approx - 0.538$ and $x = 3$ .
Check for the second solution: $f (3) = 4 \cdot \frac{1}{8} + 4.5 - 5 = 0$ .

Draw the graph for $- 2 \leq x \leq 6$ and $- 4 \leq y \leq 10$ (2 cm per 2 units on both axes), using a table of values or the GDC: $f (- 2) = 8$ , $f (0) = - 1$ , $f (6) = 4.0625$ .
The curve falls from $(- 2, 8)$ , cuts the $x$ -axis at $x \approx - 0.54$ , cuts the $y$ -axis at $(0, - 1)$ , reaches a local minimum near $(0.89; - 1.5)$ , rises and cuts the $x$ -axis again at $x = 3$ , and ends at $(6; 4.06)$ .

Graph of f(x) = 4·2^(−x) + 1.5x − 5 for −2 ≤ x ≤ 6 and −4 ≤ y ≤ 10, with the intercepts with the axes marked

(i) Since $f$ is the derivative of $g$ , $g^{'} (1) = f (1)$ : $g^{'} (1) = 4 \cdot 2^{- 1} + 1.5 - 5 = 2 + 1.5 - 5 = - 1.5.$
(ii) The tangent at $x = 1$ passes through $(1, g (1)) = (1, 3)$ and has gradient $- 1.5$ : $y - 3 = - 1.5 (x - 1) \Rightarrow y = - 1.5 x + 4.5.$

Answer: a) $f (0) = - 1$ ; b) $x \approx - 0.538$ and $x = 3$ ; c) graph with intercepts $(- 0.538; 0)$ , $(3; 0)$ and $(0; - 1)$ ; d) (i) $g^{'} (1) = - 1.5$ ; (ii) $y = - 1.5 x + 4.5$

Exercise 15 MediumIB · P1 TZ1 May 2008

Consider the function $f (x) = 2 x^{3} - 5 x^{2} + 3 x + 1$ .

a) Find $f^{'} (x)$ .

b) Write down the value of $f^{'} (2)$ .

c) Find the equation of the tangent to the curve $y = f (x)$ at the point $(2, 3)$ .

Show answer

Answer: a) $f^{'} (x) = 6 x^{2} - 10 x + 3$ ; b) $f^{'} (2) = 7$ ; c) $y = 7 x - 11$

Show working and answer

Differentiate term by term: $f^{'} (x) = 6 x^{2} - 10 x + 3.$

Substitute $x = 2$ : $f^{'} (2) = 6 \cdot 4 - 10 \cdot 2 + 3 = 24 - 20 + 3 = 7.$

Check that the point lies on the curve: $f (2) = 16 - 20 + 6 + 1 = 3$ .
The tangent has gradient $f^{'} (2) = 7$ and passes through $(2, 3)$ : $y - 3 = 7 (x - 2) \Rightarrow y = 7 x - 11.$

Answer: a) $f^{'} (x) = 6 x^{2} - 10 x + 3$ ; b) $f^{'} (2) = 7$ ; c) $y = 7 x - 11$

Exercise 16 MediumIB · P1 TZ2 May 2008

Consider the function $f (x) = \frac{1}{2} x^{3} - 2 x^{2} + 3$ .

a) Find $f^{'} (x)$ .

b) Find $f^{''} (x)$ .

c) Find the equation of the tangent to the curve of $f$ at the point $(1, 1.5)$ .

Show answer

Answer: a) $f^{'} (x) = \frac{3}{2} x^{2} - 4 x$ ; b) $f^{''} (x) = 3 x - 4$ ; c) $y = - 2.5 x + 4$

Show working and answer

Differentiate term by term: $f^{'} (x) = \frac{3}{2} x^{2} - 4 x .$

Differentiate again: $f^{''} (x) = 3 x - 4.$

Check that the point lies on the curve: $f (1) = \frac{1}{2} - 2 + 3 = \frac{3}{2} = 1.5$ .
The gradient is $f^{'} (1) = \frac{3}{2} - 4 = - \frac{5}{2} = - 2.5.$
The tangent is $y - 1.5 = - 2.5 (x - 1) \Rightarrow y = - 2.5 x + 4.$

Answer: a) $f^{'} (x) = \frac{3}{2} x^{2} - 4 x$ ; b) $f^{''} (x) = 3 x - 4$ ; c) $y = - 2.5 x + 4$

Exercise 17 MediumIB · P1 TZ2 May 2009

Consider $f : x \to x^{2} - 4$ .

a) Find $f^{'} (x)$ .

Let L be the line with equation $y = 3 x + 2$ .

b) Write down the gradient of a line parallel to L.

c) Let P be a point on the curve of $f$ . At P, the tangent to the curve is parallel to L.

Find the coordinates of P.

Show answer

Answer: a) $f^{'} (x) = 2 x$ ; b) $3$ ; c) $P (1.5; - 1.75)$

Show working and answer

Differentiate $f (x) = x^{2} - 4$ : $f^{'} (x) = 2 x .$

The line $y = 3 x + 2$ has gradient $3$ , and parallel lines have the same gradient: $3$ .

The tangent at P is parallel to L, so its gradient is $3$ : $f^{'} (x) = 2 x = 3 \Rightarrow x = \frac{3}{2} .$
Calculate the $y$ -coordinate: $f (\frac{3}{2}) = \frac{9}{4} - 4 = - \frac{7}{4} .$
Therefore $P (\frac{3}{2}, - \frac{7}{4}) = P (1.5; - 1.75)$ .

Answer: a) $f^{'} (x) = 2 x$ ; b) $3$ ; c) $P (1.5; - 1.75)$

Exercise 18 MediumIB · P2 TZ2 May 2015

Consider the function $f (x) = 0.5 x^{2} - \frac{8}{x}$ , $x \neq = 0$ .

a) Find $f (- 2)$ .

b) Find $f^{'} (x)$ .

c) Find the gradient of the graph of $f$ at $x = - 2$ .

Let $T$ be the tangent to the graph of $f$ at $x = - 2$ .

d) Write down the equation of $T$ .

e) Draw the graph of $f$ for $- 5 \leq x \leq 5$ and $- 20 \leq y \leq 20$ .

f) Draw $T$ on your sketch.

The tangent $T$ cuts the graph of $f$ at a second point, $P$ .

g) Use your GDC to find the coordinates of $P$ .

Show answer

Answer: a) $f (- 2) = 6$ ; b) $f^{'} (x) = x + \frac{8}{x ^{2}}$ ; c) $0$ ; d) $T : y = 6$ ; e) and f) graph with vertical asymptote $x = 0$ , local minimum $(- 2, 6)$ and horizontal tangent $y = 6$ ; g) $P (4, 6)$

Show working and answer

a) $f (- 2) = 0.5 \cdot 4 - \frac{8}{- 2} = 2 + 4 = 6.$

Write $f (x) = 0.5 x^{2} - 8 x^{- 1}$ and differentiate: $f^{'} (x) = x + 8 x^{- 2} = x + \frac{8}{x ^{2}} .$

c) $f^{'} (- 2) = - 2 + \frac{8}{4} = 0.$

The gradient is $0$ , so the tangent is horizontal and passes through $(- 2, 6)$ .
The equation of $T$ is $y = 6.$

e) and f)

Draw $f$ for $- 5 \leq x \leq 5$ and $- 20 \leq y \leq 20$ . The $y$ -axis ( $x = 0$ ) is a vertical asymptote.
The left branch starts at $f (- 5) = 14.1$ , falls to the local minimum $(- 2, 6)$ and rises towards $+ \infty$ near $x = 0$ .
The right branch comes from $- \infty$ near $x = 0$ , cuts the $x$ -axis at $x \approx 2.52$ and ends at $f (5) = 14.1$ .
The tangent $T$ is the horizontal line $y = 6$ , which touches the graph at $(- 2, 6)$ .

Graph of f(x) = 0.5x² − 8/x for −5 ≤ x ≤ 5 and −20 ≤ y ≤ 20, with the horizontal tangent T: y = 6 and the point P(4; 6)

Using the GDC, look for the other intersection of $y = f (x)$ with $y = 6$ : solve $0.5 x^{2} - \frac{8}{x} = 6.$
Apart from $x = - 2$ (the point of tangency), the solution is $x = 4$ . (Algebraically, multiplying by $2 x$ gives $x^{3} - 12 x - 16 = (x + 2)^{2} (x - 4) = 0$ .)
The $y$ -coordinate is $f (4) = 8 - 2 = 6$ , so $P (4, 6)$ .

Exercise 19 MediumIB · P1 TZ2 May 2014

Consider the graph of the function $f (x) = x^{3} + 2 x^{2} - 5$ .

Graph of f(x) = x³ + 2x² − 5 for −5 ≤ x ≤ 5 and −20 ≤ y ≤ 10, with tick marks at the ends of the axes

a) On the graph, label the local maximum with an A.

b) On the graph, label the local minimum with a B.

c) Write down the interval in which $f^{'} (x) < 0$ .

d) On the graph, draw accurately the tangent to the curve at $x = 1$ .

e) Write down the equation of the tangent at $x = 1$ .

Show answer

Answer: a) and b) local maximum A at $x = - \frac{4}{3}$ (point $(- 1.33; - 3.81)$ ) and local minimum B at $(0, - 5)$ ; c) $- \frac{4}{3} < x < 0$ ; d) tangent at $x = 1$ drawn through the point $(1, - 2)$ with gradient $7$ ; e) $y = 7 x - 9$

Show working and answer

a) and b)

The local extrema are where $f^{'} (x) = 0$ : $f^{'} (x) = 3 x^{2} + 4 x = x (3 x + 4) = 0 \Rightarrow x = - \frac{4}{3} or x = 0.$
The second derivative is $f^{''} (x) = 6 x + 4$ . At $x = - \frac{4}{3}$ it equals $- 4 < 0$ : local maximum. At $x = 0$ it equals $4 > 0$ : local minimum.
We calculate the $y$ -coordinates: $f (- \frac{4}{3}) = - \frac{103}{27} \approx - 3.81$ and $f (0) = - 5$ .
Label the local maximum $(- \frac{4}{3}; - 3.81) \approx (- 1.33; - 3.81)$ with A, and the local minimum $(0, - 5)$ with B.

$f$ is decreasing where $f^{'} (x) < 0$ : $x (3 x + 4) < 0.$
The parabola $3 x^{2} + 4 x$ is negative between its roots, so $- \frac{4}{3} < x < 0.$

To draw the tangent at $x = 1$ , use the point $(1, f (1)) = (1, - 2)$ and the gradient $f^{'} (1) = 7$ (calculated in part (e)): a line rising steeply to the right, touching the curve only at that point.

Graph of f(x) = x³ + 2x² − 5 with the local maximum A, the local minimum B and the tangent at x = 1 (gradient 7, through the point (1; −2))

Calculate the point and the gradient: $f (1) = 1 + 2 - 5 = - 2 f^{'} (1) = 3 + 4 = 7.$
The tangent is $y + 2 = 7 (x - 1) \Rightarrow y = 7 x - 9.$

Exercise 20 MediumIB · P2 TZ1 May 2013

The graph of the function $f (x) = \frac{14}{x} + x - 6$ , for $1 \leq x \leq 7$ , is shown below.

Graph of f(x) between x = 1 and x = 7, with no numerical scale: it starts at A at the top left, falls to a local minimum near the middle of the interval and rises to B, on the right and slightly below A

a) Calculate $f (1)$ .

b) Calculate $f^{'} (x)$ .

c) Use your answer to part (b) to show that the $x$ -coordinate of the local minimum of the graph of $f$ is 3.7, correct to 2 significant figures.

d) Find the range of $f$ .

Points $A$ and $B$ lie on the graph of $f$ . The $x$ -coordinates of $A$ and $B$ are 1 and 7 respectively.

e) Write down the $y$ -coordinate of $B$ .

f) Find the gradient of the straight line passing through $A$ and $B$ .

$M$ is the midpoint of the line segment $A B$ .

g) Write down the coordinates of $M$ .

$L$ is the tangent to the graph of the function $y = f (x)$ at the point on the graph with the same $x$ -coordinate as $M$ .

h) Find the gradient of $L$ .

(i) Find the equation of $L$ . Give your answer in the form $y = m x + c$ .

Show answer

Answer: a) $9$ ; b) $f^{'} (x) = - \frac{14}{x ^{2}} + 1$ ; c) $x = 14 \approx 3.7$ ; d) $214 - 6 \leq f (x) \leq 9$ ( $\approx 1.48 \leq f (x) \leq 9$ ); e) $3$ ; f) $- 1$ ; g) $M (4, 6)$ ; h) $\frac{1}{8}$ ; i) $y = \frac{1}{8} x + 1$

Show working and answer

a) $f (1) = \frac{14}{1} + 1 - 6 = 9.$

We write $f (x) = 14 x^{- 1} + x - 6$ and differentiate: $f^{'} (x) = - \frac{14}{x ^{2}} + 1.$

A local minimum satisfies $f^{'} (x) = 0$ : $1 - \frac{14}{x ^{2}} = 0 \Rightarrow x^{2} = 14 \Rightarrow x = 14 = 3.7416 \dots$
Only the positive root is valid (the domain is $1 \leq x \leq 7$ ). Rounding to 2 significant figures, $x = 3.7$ .
It is a minimum: $f^{'}$ is negative before $14$ and positive after it.

The minimum value is $f (14) = \frac{14}{14} + 14 - 6 = 214 - 6 \approx 1.48$ .
At the endpoints: $f (1) = 9$ and $f (7) = 2 + 7 - 6 = 3$ . The maximum value is $9$ .
Range: $214 - 6 \leq f (x) \leq 9 (\approx 1.48 \leq f (x) \leq 9) .$

e) $y_{B} = f (7) = 2 + 7 - 6 = 3.$

f) $m_{A B} = \frac{3 - 9}{7 - 1} = - 1.$

g) $M = (\frac{1 + 7}{2}, \frac{9 + 3}{2}) = (4, 6) .$

The tangent $L$ touches the curve at $x = 4$ , so its gradient is $f^{'} (4)$ : $f^{'} (4) = 1 - \frac{14}{16} = \frac{1}{8} = 0.125.$

The point of tangency lies on the curve: $f (4) = \frac{14}{4} + 4 - 6 = \frac{3}{2}$ .
Equation of $L$ : $y - \frac{3}{2} = \frac{1}{8} (x - 4) \Rightarrow y = \frac{1}{8} x + 1.$

Exercise 21 MediumIB · P2 TZ0 November 2013

Consider the function $f (x) = \frac{3}{4} x^{4} - x^{3} - 9 x^{2} + 20$ .

a) Find $f (- 2)$ .

b) Find $f^{'} (x)$ .

The graph of the function $f (x)$ has a local minimum at the point where $x = - 2$ .

c) Using your answer to part (b), verify that there is a second local minimum at $x = 3$ .

d) Sketch the graph of $f (x)$ for $- 5 \leq x \leq 5$ and $- 40 \leq y \leq 50$ . State the coordinates of the $y$ -intercept on your sketch.

e) Write down the coordinates of the local maximum.

Let $T$ be the tangent to the graph of $f (x)$ at the point $(2, - 12)$ .

f) Find the gradient of $T$ .

The line $L$ passes through the point $(2, - 12)$ and is perpendicular to $T$ .

$L$ has equation $x + b y + c = 0$ , where $b$ and $c \in Z$ .

g) Find

(i) the gradient of $L$ ;

(ii) the values of $b$ and $c$ .

Show answer

Answer: a) $4$ ; b) $f^{'} (x) = 3 x^{3} - 3 x^{2} - 18 x$ ; c) verified ( $f^{'} (3) = 0$ with a change of sign from $-$ to $+$ ); d) graph with $y$ -intercept at $(0, 20)$ ; e) $(0, 20)$ ; f) $- 24$ ; g) (i) $\frac{1}{24}$ ; (ii) $b = - 24$ , $c = - 290$

Show working and answer

a) $f (- 2) = \frac{3}{4} (- 2)^{4} - (- 2)^{3} - 9 (- 2)^{2} + 20 = 12 + 8 - 36 + 20 = 4.$

b) $f^{'} (x) = 3 x^{3} - 3 x^{2} - 18 x .$

Factorising: $f^{'} (x) = 3 x (x^{2} - x - 6) = 3 x (x - 3) (x + 2)$ .
It is zero at $x = - 2$ , $x = 0$ and $x = 3$ . We study the sign of $f^{'}$ :

for $x < - 2$ : negative;
for $- 2 < x < 0$ : positive;
for $0 < x < 3$ : negative;
for $x > 3$ : positive.

At $x = 3$ the derivative is zero and changes from negative to positive, so there is a second local minimum. Its value is $f (3) = - 27.25$ .

We evaluate key points: $f (0) = 20$ (the $y$ -intercept), a minimum at $(- 2, 4)$ , a maximum at $(0, 20)$ and a minimum at $(3, - 27.25)$ .
The graph falls from the left to $(- 2, 4)$ , rises to $(0, 20)$ , falls to $(3, - 27.25)$ and then rises again. It crosses the $x$ -axis near $x = 1.5$ and $x = 3.9$ .

Graph of f(x) = 3/4 x⁴ − x³ − 9x² + 20 in the window −5 ≤ x ≤ 5, −40 ≤ y ≤ 50: local minimum at (−2; 4), local maximum at (0; 20), local minimum at (3; −27.25) and y-intercept at (0; 20)

Between the two minima there is a local maximum, located at $x = 0$ (where $f^{'}$ changes from positive to negative): $f (0) = 20$ .
Local maximum: $(0, 20)$ .

f) $f^{'} (2) = 3 \cdot 8 - 3 \cdot 4 - 18 \cdot 2 = 24 - 12 - 36 = - 24.$

(i) $L$ is perpendicular to $T$ : $m_{L} = - \frac{1}{- 24} = \frac{1}{24}$ .
(ii) In $x + b y + c = 0$ the gradient is $- \frac{1}{b}$ . Equating: $- \frac{1}{b} = \frac{1}{24} \Rightarrow b = - 24$ .
Since $L$ passes through $(2, - 12)$ : $2 + (- 24) (- 12) + c = 0 \Rightarrow 2 + 288 + c = 0 \Rightarrow c = - 290.$

Exercise 22 MediumIB · P2 TZ0 May 2006

The function $g (x)$ is defined by $g (x) = \frac{1}{8} x^{4} + \frac{9}{4} x^{2} - 5 x + 7$ , $x \geq 0$ .

a) Find $g (2)$ .

b) Calculate $g^{'} (x)$ .

The graph of the function $y = g (x)$ has a tangent $T_{1}$ at the point where $x = 2$ .

c) (i) Show that the gradient of $T_{1}$ is 8.

(ii) Find the equation of $T_{1}$ . Write the equation in the form $y = m x + c$ .

d) The graph has another tangent, $T_{2}$ , at the point $(1, \frac{35}{8})$ . $T_{2}$ has gradient zero. Write down the equation of $T_{2}$ .

e) (i) Sketch the graph of $y = g (x)$ in the region $0 \leq x \leq 3$ , $0 \leq y \leq 22$ .

(ii) Add the tangents $T_{1}$ and $T_{2}$ to your sketch, in the correct positions.

Show answer

Answer: a) $8$ ; b) $g^{'} (x) = \frac{1}{2} x^{3} + \frac{9}{2} x - 5$ ; c) (i) shown, gradient $8$ ; (ii) $T_{1} : y = 8 x - 8$ ; d) $T_{2} : y = \frac{35}{8}$ ; e) graph with $T_{1}$ and $T_{2}$ drawn

Show working and answer

a) $g (2) = \frac{1}{8} \cdot 16 + \frac{9}{4} \cdot 4 - 10 + 7 = 2 + 9 - 10 + 7 = 8.$

b) $g^{'} (x) = \frac{1}{2} x^{3} + \frac{9}{2} x - 5.$

(i) $g^{'} (2) = \frac{1}{2} \cdot 8 + \frac{9}{2} \cdot 2 - 5 = 4 + 9 - 5 = 8$ .
(ii) The tangent passes through $(2, 8)$ with gradient $8$ : $y - 8 = 8 (x - 2) \Rightarrow y = 8 x - 8.$

Gradient zero at $x = 1$ (check: $g^{'} (1) = \frac{1}{2} + \frac{9}{2} - 5 = 0$ ) and $g (1) = \frac{1}{8} + \frac{9}{4} - 5 + 7 = \frac{35}{8}$ .
Horizontal tangent through $(1, \frac{35}{8})$ : $T_{2} : y = \frac{35}{8} = 4.375.$

(i) Points on the curve: $g (0) = 7$ , minimum at $(1, \frac{35}{8})$ , $g (2) = 8$ and $g (3) \approx 22.4$ (the curve falls to $x = 1$ and then rises ever more steeply).
(ii) $T_{1}$ is the line of gradient $8$ through $(2, 8)$ (it meets the $x$ -axis at $(1, 0)$ ); $T_{2}$ is the horizontal line $y = 4.375$ , which touches the curve at its minimum.

Graph of y = g(x) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 23, with a local minimum at (1; 4.38), the tangent T1 of gradient 8 passing through (2; 8) and meeting the x-axis at (1; 0), and the horizontal tangent T2, y = 4.38, touching the curve at its minimum

Exercise 23 MediumIB · P1 TZ2 May 2011

The diagram shows the graphs of the functions $f (x) = \frac{1}{4} x^{2} - 2$ and $g (x) = x$ .

Graphs of f(x) = x²/4 − 2 (parabola with vertex (0; −2)) and g(x) = x (line through the origin) on the same axes, with x from −6 to 6 and y from −6 to 4, and a grid

a) Differentiate $f (x)$ with respect to $x$ .

b) Differentiate $g (x)$ with respect to $x$ .

c) Calculate the value of $x$ for which the gradients of the two graphs are equal.

d) Draw the tangent to the parabola at the point with the $x$ -value found in part (c).

Show answer

Answer: a) $f^{'} (x) = \frac{x}{2}$ ; b) $g^{'} (x) = 1$ ; c) $x = 2$ ; d) tangent $y = x - 3$ at the point $(2, - 1)$

Show working and answer

a) $f^{'} (x) = \frac{1}{4} \cdot 2 x = \frac{x}{2} .$

b) $g^{'} (x) = 1.$

Equate the gradients: $\frac{x}{2} = 1$ .
Hence $x = 2$ .

Point on the parabola: $f (2) = \frac{4}{4} - 2 = - 1$ , that is, $(2, - 1)$ .
Gradient: $f^{'} (2) = 1$ . The tangent is $y + 1 = 1 \cdot (x - 2) \Rightarrow y = x - 3.$
Draw the line $y = x - 3$ : it passes through $(2, - 1)$ , $(3, 0)$ and $(0, - 3)$ , and is parallel to $g$ .

Same axes as the question: parabola f(x) = x²/4 − 2 and line g(x) = x, with the tangent line at (2; −1), of gradient 1 and equation y = x − 3, parallel to g; the point of tangency (2; −1) is marked

Answer: a) $f^{'} (x) = \frac{x}{2}$ ; b) $g^{'} (x) = 1$ ; c) $x = 2$ ; d) tangent $y = x - 3$ at the point $(2, - 1)$

Exercise 24 MediumIB · P1 TZ0 May 2007

The figure shows the graphs of the functions $f_{1} (x) = x$ and $f_{2} (x) = 5 - x^{2}$ .

Graphs of f1(x)=x and f2(x)=5−x² on the same axes, with grid

a) (i) Differentiate $f_{1} (x)$ with respect to $x$ .

(ii) Differentiate $f_{2} (x)$ with respect to $x$ .

b) Calculate the value of $x$ for which the two graphs have the same gradient.

c) Draw accurately on the figure the tangent to the curved graph for this value of $x$ , clearly showing the property from part (b).

Show answer

Answer: a) (i) $f_{1}^{'} (x) = 1$ ; (ii) $f_{2}^{'} (x) = - 2 x$ ; b) $x = - \frac{1}{2}$ ; c) tangent $y = x + \frac{21}{4}$ , parallel to $f_{1}$ (figure)

Show working and answer

(i) $f_{1}^{'} (x) = 1$ .
(ii) $f_{2}^{'} (x) = - 2 x$ .

We equate the gradients: $1 = - 2 x \Rightarrow x = - \frac{1}{2} .$

At $x = - \frac{1}{2}$ : $f_{2} (- \frac{1}{2}) = 5 - \frac{1}{4} = \frac{19}{4} = 4.75$ , and the gradient is $1$ .
The tangent to the parabola at $(- 0.5; 4.75)$ is $y - \frac{19}{4} = 1 \cdot (x + \frac{1}{2}) \Rightarrow y = x + \frac{21}{4} .$
This line is drawn, parallel to $f_{1}$ (same gradient $1$ ), touching the parabola only at that point.

Parabola f2(x)=5−x² and line f1(x)=x on the same axes, with the tangent to f2 at (−0.5; 4.75) of gradient 1, parallel to f1, drawn as a dashed line

Answer: a) (i) $f_{1}^{'} (x) = 1$ ; (ii) $f_{2}^{'} (x) = - 2 x$ ; b) $x = - \frac{1}{2}$ ; c) tangent $y = x + \frac{21}{4}$ , parallel to $f_{1}$ (figure)

Exercise 25 HardIB · P2 TZ0 November 2019

The graph of the quadratic function $f (x) = \frac{1}{2} (x - 2) (x + 8)$ cuts the $y$ -axis at $(0, c)$ .

The vertex of the function is $(- 3; - 12.5)$ .

a) Find the value of $c$ .

b) Write down the equation of the axis of symmetry of the graph.

The equation $f (x) = 12$ has two solutions. The first solution is $x = - 10$ .

c) Use the symmetry of the graph to show that the second solution is $x = 4$ .

d) Write down the $x$ -intercepts of the graph.

e) On graph paper, draw the graph of $y = f (x)$ accurately for $- 10 \leq x \leq 4$ and $- 14 \leq y \leq 14$ . Use a scale of 1 cm to represent 1 unit on the $x$ -axis and 1 cm to represent 2 units on the $y$ -axis.

Let $T$ be the tangent at $x = - 3$ .

f) (i) Write down the equation of $T$ .

(ii) On the same graph, draw the tangent $T$ accurately.

g) Given that $f (a) = 5.5$ and $f^{'} (a) = - 6$ , state whether the function $f$ is increasing or decreasing at $x = a$ . Give a reason to justify your answer.

Show answer

Answer: a) $c = - 8$ ; b) $x = - 3$ ; c) $x = 4$ (shown); d) $(2, 0)$ and $(- 8, 0)$ ; e) parabola with vertex $(- 3; - 12.5)$ and points $(- 10, 12)$ , $(- 8, 0)$ , $(0, - 8)$ , $(2, 0)$ , $(4, 12)$ ; f) (i) $T : y = - 12.5$ , (ii) horizontal line through the vertex; g) decreasing at $x = a$ , because $f^{'} (a) < 0$

Show working and answer

The $y$ -intercept is found by setting $x = 0$ : $c = f (0) = \frac{1}{2} (0 - 2) (0 + 8) = \frac{1}{2} (- 16) = - 8.$

The axis of symmetry is the vertical line through the vertex: $x = - 3$ .

The two points with the same image are the same distance from the axis of symmetry, so $- 3$ is the midpoint of the two solutions.
With $x_{1} = - 10$ : $\frac{- 10 + x _{2}}{2} = - 3 \Rightarrow x_{2} = - 6 + 10 = 4.$

$f (x) = 0$ when $x = 2$ or $x = - 8$ .
The $x$ -intercepts are $(2, 0)$ and $(- 8, 0)$ .

We choose points: $(- 10, 12)$ , $(- 8, 0)$ , vertex $(- 3; - 12.5)$ , $(0, - 8)$ , $(2, 0)$ and $(4, 12)$ , and join them with a parabola opening upwards.

Graph of f(x) = ½(x − 2)(x + 8) on graph paper, parabola with vertex (−3; −12.5), x-intercepts at (−8, 0) and (2, 0), y-intercept at (0, −8), and the horizontal tangent line y = −12.5 at the vertex

(i) We differentiate: $f^{'} (x) = \frac{1}{2} (2 x + 6) = x + 3$ , and $f^{'} (- 3) = 0$ . The tangent at the vertex is horizontal and passes through $y = - 12.5$ : $T : y = - 12.5.$
(ii) On the graph, $T$ is the horizontal line that touches the parabola at its vertex $(- 3; - 12.5)$ .

Since $f^{'} (a) = - 6 < 0$ , the gradient of the curve at $x = a$ is negative.
Therefore, $f$ is decreasing at $x = a$ .

Exercise 26 HardIB · P2 TZ1 May 2017

Consider the function $g (x) = x^{3} + k x^{2} - 15 x + 5$ .

a) Find $g^{'} (x)$ .

The tangent to the graph of $y = g (x)$ at $x = 2$ is parallel to the line $y = 21 x + 7$ .

b) (i) Show that $k = 6$ .

(ii) Find the equation of the tangent to the graph of $y = g (x)$ at $x = 2$ . Give your answer in the form $y = m x + c$ .

c) Use your answer to part (a) and the value of $k$ to find the $x$ -coordinates of the stationary points of the graph of $y = g (x)$ .

d) (i) Find $g^{'} (- 1)$ .

(ii) Using the above, justify that $g$ is decreasing at $x = - 1$ .

e) Find the $y$ -coordinate of the local minimum.

Show answer

Answer: a) $g^{'} (x) = 3 x^{2} + 2 k x - 15$ ; b) (i) $k = 6$ (shown); (ii) $y = 21 x - 35$ ; c) $x = 1$ and $x = - 5$ ; d) (i) $- 24$ ; (ii) $g^{'} (- 1) < 0$ , decreasing; e) $y = - 3$

Show working and answer

$g^{'} (x) = 3 x^{2} + 2 k x - 15$ .

(i) The gradient of the tangent at $x = 2$ is $g^{'} (2) = 12 + 4 k - 15 = 4 k - 3$ . It is parallel to $y = 21 x + 7$ , which has gradient $21$ : $4 k - 3 = 21 \Rightarrow k = 6.$
(ii) With $k = 6$ : $g (2) = 8 + 24 - 30 + 5 = 7$ and $g^{'} (2) = 21$ .
Equation of the tangent: $y - 7 = 21 (x - 2) \Rightarrow y = 21 x - 35.$

With $k = 6$ : $g^{'} (x) = 3 x^{2} + 12 x - 15 = 3 (x - 1) (x + 5)$ .
The stationary points satisfy $g^{'} (x) = 0$ : $x = 1$ and $x = - 5$ .

(i) $g^{'} (- 1) = 3 - 12 - 15 = - 24$ .
(ii) Since $g^{'} (- 1) = - 24 < 0$ , the gradient at $x = - 1$ is negative and $g$ is decreasing at $x = - 1$ .

$g^{'}$ changes from negative to positive at $x = 1$ (it is negative at $x = - 1$ and positive, for example, at $x = 2$ ), so there is a local minimum there.
Its $y$ -coordinate is $g (1) = 1 + 6 - 15 + 5 = - 3$ .

Answer: a) $g^{'} (x) = 3 x^{2} + 2 k x - 15$ ; b) (i) $k = 6$ (shown); (ii) $y = 21 x - 35$ ; c) $x = 1$ and $x = - 5$ ; d) (i) $- 24$ ; (ii) $g^{'} (- 1) < 0$ , decreasing; e) $y = - 3$

Exercise 27 HardIB · P2 TZ0 November 2007

The diagram below shows the graph of a straight line $L$ passing through $(1, 1)$ and $(2, 3)$ , and the graph $P$ of the function $f (x) = x^{2} - 3 x - 4$ .

Line L with positive gradient passing through (1, 1) and (2, 3), and parabola P with equation y = x² − 3x − 4, with its vertex below the x-axis to the right of the y-axis; they intersect to the left of the y-axis, below the origin

a) Find the gradient of the line $L$ .

b) Differentiate $f (x)$ .

c) Find the coordinates of the point where the tangent to $P$ is parallel to the line $L$ .

d) Find the coordinates of the point where the tangent to $P$ is perpendicular to the line $L$ .

e) Find

(i) the gradient of the tangent to $P$ at the point with coordinates $(2, - 6)$ .

(ii) the equation of the tangent to $P$ at this point.

f) State the equation of the axis of symmetry of $P$ .

g) Find the coordinates of the vertex of $P$ and state the gradient of the curve at this point.

Show answer

Answer: a) $2$ ; b) $f^{'} (x) = 2 x - 3$ ; c) $(2.5, - 5.25)$ ; d) $(1.25, - 6.19)$ ; e) (i) $1$ ; (ii) $y = x - 8$ ; f) $x = 1.5$ ; g) vertex $(1.5, - 6.25)$ and gradient $0$

Show working and answer

Gradient of $L$ between $(1, 1)$ and $(2, 3)$ : $m = \frac{3 - 1}{2 - 1} = 2.$

$f^{'} (x) = 2 x - 3$ .

A tangent parallel to $L$ has gradient $2$ : $f^{'} (x) = 2$ .
$2 x - 3 = 2 \Rightarrow x = \frac{5}{2}$ .
$f (\frac{5}{2}) = \frac{25}{4} - \frac{15}{2} - 4 = - \frac{21}{4} = - 5.25$ .
The point is $(2.5, - 5.25)$ .

A tangent perpendicular to $L$ has gradient $- \frac{1}{2}$ : $2 x - 3 = - \frac{1}{2}$ , so $x = \frac{5}{4}$ .
$f (\frac{5}{4}) = \frac{25}{16} - \frac{15}{4} - 4 = - \frac{99}{16} \approx - 6.19$ .
The point is $(1.25, - 6.19)$ .

(i) $f^{'} (2) = 4 - 3 = 1$ .
(ii) The tangent passes through $(2, - 6)$ with gradient $1$ : $y + 6 = 1 \cdot (x - 2) \Rightarrow y = x - 8.$

The axis of symmetry passes through the vertex, where $f^{'} (x) = 0$ : $2 x - 3 = 0$ , so $x = 1.5$ .

The $x$ -coordinate of the vertex is $x = 1.5$ and $f (1.5) = \frac{9}{4} - \frac{9}{2} - 4 = - \frac{25}{4} = - 6.25$ . The vertex is $(1.5, - 6.25)$ .
At the vertex the tangent is horizontal: the gradient is $f^{'} (1.5) = 0$ .

Answer: a) $2$ ; b) $f^{'} (x) = 2 x - 3$ ; c) $(2.5, - 5.25)$ ; d) $(1.25, - 6.19)$ ; e) (i) $1$ ; (ii) $y = x - 8$ ; f) $x = 1.5$ ; g) vertex $(1.5, - 6.25)$ and gradient $0$

Exercise 28 HardIB · P2 TZ0 November 2012

Consider the function $g (x) = b x - 3 + \frac{1}{x ^{2}}$ , $x \neq = 0$ .

a) Write down the equation of the vertical asymptote of the graph of $y = g (x)$ .

b) Write down $g^{'} (x)$ .

The line $T$ is the tangent to the graph of $y = g (x)$ at the point where $x = 1$ . The gradient of $T$ is $3$ .

c) Show that $b = 5$ .

d) Find the equation of $T$ .

e) Using your GDC, find the coordinates of the point where the graph of $y = g (x)$ meets the $x$ -axis.

f) (i) Draw the graph of $y = g (x)$ for $- 2 \leq x \leq 5$ and $- 15 \leq y \leq 25$ , clearly indicating your answer to part (e).

(ii) Draw the line $T$ on your graph.

g) Using your GDC, find the coordinates of the local minimum point of $y = g (x)$ .

h) Write down the interval for which $g (x)$ is increasing in the domain $0 < x < 5$ .

Show answer

Answer: a) $x = 0$ ; b) $g^{'} (x) = b - \frac{2}{x ^{3}}$ ; c) shown; d) $y = 3 x$ ; e) $(- 0.439; 0)$ ; f) graph with $T$ ; g) $(0.737; 2.53)$ ; h) $0.737 < x < 5$

Show working and answer

The denominator $x^{2}$ is zero at $x = 0$ , and $g (x) \to + \infty$ as $x \to 0$ .
Vertical asymptote: $x = 0$ .

Write $g (x) = b x - 3 + x^{- 2}$ and differentiate: $g^{'} (x) = b - \frac{2}{x ^{3}} .$

The gradient of $T$ at $x = 1$ is $g^{'} (1) = b - 2$ .
Since this gradient equals $3$ : $b - 2 = 3$ , so $b = 5$ .

With $b = 5$ : $g (1) = 5 - 3 + 1 = 3$ , so the point of tangency is $(1, 3)$ .
Tangent: $y - 3 = 3 (x - 1) \Rightarrow y = 3 x .$

Using the GDC, solve $5 x - 3 + \frac{1}{x ^{2}} = 0$ (equivalent to $5 x^{3} - 3 x^{2} + 1 = 0$ , which has only one real root).
$x \approx - 0.439$ . The point is $(- 0.439; 0)$ .

(i) There are two branches, separated by the asymptote $x = 0$ . Both rise to $+ \infty$ as they approach $x = 0$ . The left branch starts at $g (- 2) = - 12.75$ and meets the $x$ -axis at $(- 0.439; 0)$ . The right branch falls to a local minimum and then rises, passing through $(1, 3)$ , $(2; 7.25)$ and $(5; 22.04)$ .
(ii) The line $T$ : $y = 3 x$ passes through the origin and touches the graph at $(1, 3)$ .

Graph of y = 5x − 3 + 1/x² for −2 ≤ x ≤ 5 and −15 ≤ y ≤ 25, with grid and the vertical asymptote at x = 0 dashed; it crosses the x-axis at (−0.439; 0), has a local minimum at (0.737; 2.53), and the line T, y = 3x, is tangent at (1; 3)

Local minimum: $g^{'} (x) = 0 \Rightarrow 5 = \frac{2}{x ^{3}} \Rightarrow x^{3} = \frac{2}{5}$ .
Using the GDC (or directly), $x = (\frac{2}{5})^{1/3} \approx 0.737$ .
$g (0.737) = 5 (0.737) - 3 + \frac{1}{0.73 7 ^{2}} \approx 2.53$ . Minimum point: $(0.737; 2.53)$ .

$g^{'} (x) > 0 ⟺ 5 > \frac{2}{x ^{3}} ⟺ x^{3} > \frac{2}{5} ⟺ x > 0.737$ (for $x > 0$ ).
$g$ is increasing for $0.737 < x < 5$ .

Answer: a) $x = 0$ ; b) $g^{'} (x) = b - \frac{2}{x ^{3}}$ ; c) shown; d) $y = 3 x$ ; e) $(- 0.439; 0)$ ; f) graph with $T$ ; g) $(0.737; 2.53)$ ; h) $0.737 < x < 5$

Exercise 29 HardIB · P2 TZ0 November 2011

The diagram shows part of the graph of $f (x) = x^{2} - 2 x + \frac{9}{x}$ , where $x \neq = 0$ .

Part of the graph of f(x) = x² − 2x + 9/x with x from −3 to 4 and y from −10 to 10, with grid and vertical asymptote x = 0, the curve not drawn at x = 0: the left branch meets the x-axis at (−1.58; 0); the right branch has a local minimum at (2.06; 4.49)

a) Write

(i) the equation of the vertical asymptote to the graph of $y = f (x)$ ;

(ii) the solution of the equation $f (x) = 0$ ;

(iii) the coordinates of the local minimum point.

b) Find $f^{'} (x)$ .

c) Show that $f^{'} (x)$ can be written as $f^{'} (x) = \frac{2 x ^{3} - 2 x ^{2} - 9}{x ^{2}}$ .

d) Find the gradient of the tangent to $y = f (x)$ at the point $A (1, 8)$ . The line $L$ passes through the point $A$ and is perpendicular to the tangent at $A$ .

e) Write down the gradient of $L$ .

f) Find the equation of $L$ . Give your answer in the form $y = m x + c$ .

$L$ also cuts the graph of $y = f (x)$ at the points $B$ and $C$ .

g) Write down the $x$ -coordinate of $B$ and of $C$ .

Show answer

Answer: a) (i) $x = 0$ ; (ii) $x \approx - 1.58$ ; (iii) $(2.06; 4.49)$ ; b) $f^{'} (x) = 2 x - 2 - \frac{9}{x ^{2}}$ ; c) shown; d) $- 9$ ; e) $\frac{1}{9}$ ; f) $y = \frac{1}{9} x + \frac{71}{9}$ ; g) $x \approx - 2.50$ and $x \approx 3.61$

Show working and answer

(i) The function is not defined at $x = 0$ (the denominator is zero), and the graph shoots off towards $\pm \infty$ near that value. The vertical asymptote is $x = 0$ .
(ii) Solve $f (x) = 0$ using the GDC: the graph crosses the $x$ -axis only once, at $x \approx - 1.58$ .
(iii) The local minimum of the right branch, read from the GDC, is the point $(2.06; 4.49)$ .

Write $f (x) = x^{2} - 2 x + 9 x^{- 1}$ and differentiate term by term: $f^{'} (x) = 2 x - 2 - 9 x^{- 2} = 2 x - 2 - \frac{9}{x ^{2}} .$

Put everything over the denominator $x^{2}$ : $f^{'} (x) = \frac{( 2 x - 2 ) x ^{2}}{x ^{2}} - \frac{9}{x ^{2}} = \frac{2 x ^{3} - 2 x ^{2} - 9}{x ^{2}} .$

The gradient of the tangent at $A$ is $f^{'} (1)$ : $f^{'} (1) = \frac{2 - 2 - 9}{1} = - 9.$

The perpendicular has the negative reciprocal gradient: $m_{L} = - \frac{1}{- 9} = \frac{1}{9}$ .

$L$ passes through $A (1, 8)$ with gradient $\frac{1}{9}$ : $y - 8 = \frac{1}{9} (x - 1) \Rightarrow y = \frac{1}{9} x + \frac{71}{9} .$

Equate the line and the function using the GDC: $x^{2} - 2 x + \frac{9}{x} = \frac{x}{9} + \frac{71}{9}$ .
Apart from $x = 1$ (the point $A$ ), the other two solutions are $x \approx - 2.50$ and $x \approx 3.61$ .

Exercise 30 HardIB · P2 TZ0 November 2008

Consider the curve $y = x^{3} + \frac{3}{2} x^{2} - 6 x - 2$ .

a) (i) Write down the value of $y$ when $x$ is $2$ .

(ii) Write down the coordinates of the point where the curve crosses the $y$ -axis.

b) Sketch the curve for $- 4 \leq x \leq 3$ and $- 10 \leq y \leq 10$ . Clearly indicate the information obtained in part (a).

c) Find $\frac{d y}{d x}$ .

d) Let $L_{1}$ be the tangent to the curve at the point where $x = 2$ .

Let $L_{2}$ be a tangent to the curve, parallel to $L_{1}$ .

(i) Verify that the gradient of $L_{1}$ is equal to $12$ .

(ii) Find the $x$ -coordinate of the point at which $L_{2}$ and the curve meet.

(iii) On the diagram you drew in part (b), sketch approximately $L_{1}$ and $L_{2}$ and label both lines.

e) It is known that $\frac{d y}{d x} > 0$ for $x < - 2$ and $x > b$ , where $b$ is positive.

(i) Using the GDC or any other alternative method, find the value of $b$ .

(ii) Describe the behaviour of the curve in the interval $- 2 < x < b$ .

(iii) Write down the equation of the tangent to the curve at $x = - 2$ .

Show answer

Answer: a) (i) $0$ ; (ii) $(0, - 2)$ ; b) graph (figure); c) $\frac{d y}{d x} = 3 x^{2} + 3 x - 6$ ; d) (i) $12$ ; (ii) $x = - 3$ ; (iii) $L_{1}$ : $y = 12 x - 24$ and $L_{2}$ : $y = 12 x + 38.5$ drawn; e) (i) $b = 1$ ; (ii) the curve is decreasing; (iii) $y = 8$

Show working and answer

(i) $y (2) = 8 + \frac{3}{2} \cdot 4 - 12 - 2 = 8 + 6 - 12 - 2 = 0$ .
(ii) With $x = 0$ : $y = - 2$ , so the point is $(0, - 2)$ .

The curve is a cubic that rises from below on the left, has a local maximum at $(- 2; 8)$ , a local minimum at $(1; - 5.5)$ and rises again. It crosses the $y$ -axis at $(0, - 2)$ and the $x$ -axis at $(2, 0)$ (part (a)), and also at $x \approx - 3.19$ and $x \approx - 0.314$ .

Graph of y=x³+1.5x²−6x−2 for −4 ≤ x ≤ 3 and −10 ≤ y ≤ 10, with the points (0; −2) and (2; 0) and the tangents L1 and L2

c) $\frac{d y}{d x} = 3 x^{2} + 3 x - 6.$

(i) At $x = 2$ : $\frac{d y}{d x} = 12 + 6 - 6 = 12$ , so the gradient of $L_{1}$ is $12$ .
(ii) $L_{2}$ is parallel to $L_{1}$ , so its gradient is also $12$ : $3 x^{2} + 3 x - 6 = 12 \Rightarrow x^{2} + x - 6 = 0 \Rightarrow (x - 2) (x + 3) = 0.$ The solution $x = 2$ is the point of $L_{1}$ ; $L_{2}$ touches the curve at $x = - 3$ .
(iii) $L_{1}$ passes through $(2, 0)$ : $y = 12 x - 24$ . $L_{2}$ passes through $(- 3; 2.5)$ (since $y (- 3) = - 27 + 13.5 + 18 - 2 = 2.5$ ): $y = 12 x + 38.5$ . They are drawn as two parallel lines tangent to the curve and labelled $L_{1}$ and $L_{2}$ (see the figure).

(i) Using the GDC (or by factorising, $\frac{d y}{d x} = 3 (x + 2) (x - 1)$ ), the derivative is zero at $x = - 2$ and $x = 1$ , and is positive outside $[- 2, 1]$ . Hence $b = 1$ .
(ii) For $- 2 < x < 1$ the derivative is negative: the curve is decreasing (from the local maximum $(- 2, 8)$ to the local minimum $(1, - 5.5)$ ).
(iii) At $x = - 2$ : $y = - 8 + 6 + 12 - 2 = 8$ and $\frac{d y}{d x} = 12 - 6 - 6 = 0$ , so the tangent is horizontal: $y = 8.$

Increasing and decreasing, maxima and minima

Exercise 31 MediumIB · P2 TZ2 May 2018

Consider the curve $y = 2 x^{3} - 9 x^{2} + 12 x + 2$ , for $- 1 < x < 3$ .

a) Sketch the curve approximately for $- 1 < x < 3$ , $- 2 < y < 12$ .

b) A teacher asks students to make some observations about the curve.

Three students gave the following responses.

Nadia said "The point where the curve cuts the $x$ -axis lies between $- 1$ and zero".

Rick said "The curve is decreasing for $x < 1$ ".

Paula said "The gradient of the curve is less than zero between $x = 1$ and $x = 2$ ".

State the name of the student who made an incorrect observation.

c) Find the value of $y$ for $x = 1$ .

d) Find $\frac{d y}{d x}$ .

e) Show that the stationary points of the curve are at $x = 1$ and $x = 2$ .

f) Given that $2 x^{3} - 9 x^{2} + 12 x + 2 = k$ has three solutions, find the possible values of $k$ .

Show answer

Answer: a) graph with maximum $(1, 7)$ and minimum $(2, 6)$ ; b) Rick; c) $y = 7$ ; d) $\frac{d y}{d x} = 6 x^{2} - 18 x + 12$ ; e) shown ( $x = 1$ and $x = 2$ ); f) $6 < k < 7$

Show working and answer

Using the GDC, draw the curve in the window $- 1 < x < 3$ , $- 2 < y < 12$ : it rises to a local maximum at $(1, 7)$ , falls to a local minimum at $(2, 6)$ and rises again. It cuts the $y$ -axis at $(0, 2)$ .

Graph of y = 2x³ − 9x² + 12x + 2 for −1 < x < 3 and −2 < y < 12, with a local maximum at (1, 7) and a local minimum at (2, 6); the axes are marked from −1 to 3 in x and from 2 to 12 in y

Nadia: the $x$ -intercept lies between $- 1$ and $0$ (the curve goes from negative to positive values, since $y (- 1) = - 21$ and $y (0) = 2$ ). She is correct.
Rick: the curve rises up to $x = 1$ , that is, it is increasing for $x < 1$ (for example, the gradient at $x = 0$ is $12 > 0$ ). What he says is incorrect.
Paula: between $x = 1$ and $x = 2$ the curve falls, so the gradient is negative. She is correct.
The student who is wrong is Rick.

$y = 2 - 9 + 12 + 2 = 7$ .

$\frac{d y}{d x} = 6 x^{2} - 18 x + 12$ .

Stationary points satisfy $\frac{d y}{d x} = 0$ : $6 x^{2} - 18 x + 12 = 0 \Rightarrow x^{2} - 3 x + 2 = 0 \Rightarrow (x - 1) (x - 2) = 0.$
Hence $x = 1$ and $x = 2$ .

The solutions of $2 x^{3} - 9 x^{2} + 12 x + 2 = k$ are the intersections of the curve with the horizontal line $y = k$ .
The local maximum is $(1, 7)$ and the local minimum is $(2, 6)$ . There are three intersections when the line lies between the two values: $6 < k < 7.$

Answer: a) graph with maximum $(1, 7)$ and minimum $(2, 6)$ ; b) Rick; c) $y = 7$ ; d) $\frac{d y}{d x} = 6 x^{2} - 18 x + 12$ ; e) shown ( $x = 1$ and $x = 2$ ); f) $6 < k < 7$

Exercise 32 MediumIB · P1 TZ2 May 2010

The following table describes the behaviour of $f^{'} (x)$ , the derivative of $f (x)$ , on the domain $- 4 < x < 2$ .

$x$	$f^{'} (x)$
$- 4 < x < - 2$	$< 0$
$- 2$	$0$
$- 2 < x < 1$	$> 0$
$1$	$0$
$1 < x < 2$	$> 0$

The point $P (- 2, 3)$ lies on the graph of $f (x)$ .

a) State whether $f (0)$ is greater than, less than or equal to $f (- 2)$ . Give a reasoned answer.

b) Write down the equation of the tangent to the graph of $f (x)$ at the point $P$ .

c) Using the information given about $f^{'} (x)$ , state whether the point $(- 2, 3)$ is a maximum, a minimum or neither. Give a reasoned answer.

Show answer

Answer: a) $f (0) > f (- 2)$ , because $f$ is increasing on $] - 2, 1 [$ ; b) $y = 3$ ; c) minimum, because $f^{'}$ changes from negative to positive

Show working and answer

For $- 2 < x < 1$ we have $f^{'} (x) > 0$ , so $f$ is increasing on that interval.
Since $- 2 < 0 < 1$ , it follows that $f (0) > f (- 2)$ : it is greater.

At $P (- 2, 3)$ the derivative is $0$ , so the tangent is horizontal.
Its equation is $y = 3$ .

To the left of $x = - 2$ the derivative is negative ( $f$ is decreasing) and to the right it is positive ( $f$ is increasing).
Therefore $(- 2, 3)$ is a local minimum.

Answer: a) $f (0) > f (- 2)$ , because $f$ is increasing on $] - 2, 1 [$ ; b) $y = 3$ ; c) minimum, because $f^{'}$ changes from negative to positive

Exercise 33 MediumIB · P1 TZ1 May 2013

Consider the function $f (x) = a x^{3} - 3 x + 5$ with $a \neq = 0$ .

a) Calculate $f^{'} (x)$ .

b) Write down the value of $f^{'} (0)$ .

The function has a local maximum at $x = - 2$ .

c) Calculate the value of $a$ .

Show answer

Answer: a) $f^{'} (x) = 3 a x^{2} - 3$ ; b) $f^{'} (0) = - 3$ ; c) $a = \frac{1}{4}$

Show working and answer

Differentiate $f (x) = a x^{3} - 3 x + 5$ : $f^{'} (x) = 3 a x^{2} - 3.$

Substitute $x = 0$ : $f^{'} (0) = 3 a \cdot 0 - 3 = - 3.$

At a local maximum the derivative is zero, so $f^{'} (- 2) = 0$ : $3 a \cdot 4 - 3 = 0 \Rightarrow 12 a = 3 \Rightarrow a = \frac{1}{4} .$
Check: $f^{''} (x) = 6 a x$ and $f^{''} (- 2) = - 12 a = - 3 < 0$ , so there is a maximum at $x = - 2$ .

Answer: a) $f^{'} (x) = 3 a x^{2} - 3$ ; b) $f^{'} (0) = - 3$ ; c) $a = \frac{1}{4}$

Exercise 34 MediumIB · P1 TZ2 May 2014

A function is given by $f (x) = 2 x^{3} - 5 x + \frac{4}{x} + 3, - 5 \leq x \leq 10, x \neq = 0$ .

a) Write down the derivative of the function.

b) Use the GDC to find the coordinates of the local minimum of $f (x)$ within the given domain.

Show answer

Answer: a) $f^{'} (x) = 6 x^{2} - 5 - \frac{4}{x ^{2}}$ ; b) local minimum at $(1.15; 3.77)$

Show working and answer

Write $f (x) = 2 x^{3} - 5 x + 4 x^{- 1} + 3$ and differentiate: $f^{'} (x) = 6 x^{2} - 5 - \frac{4}{x ^{2}} .$

Using the GDC, draw $f$ for $- 5 \leq x \leq 10$ (with $x \neq = 0$ ) and look for the local minimum.
Algebraic check: $f^{'} (x) = 0$ gives $6 x^{4} - 5 x^{2} - 4 = 0$ . With $u = x^{2}$ , $6 u^{2} - 5 u - 4 = 0$ , from which $u = \frac{4}{3}$ (the other solution, $u = - \frac{1}{2}$ , is not valid).
For $x > 0$ we get $x = \frac{4}{3} = \frac{2 3}{3} \approx 1.15$ . It is a minimum because $f^{''} (x) = 12 x + \frac{8}{x ^{3}} > 0$ for $x > 0$ .
Its $y$ -coordinate is $f (1.155) \approx 3.77$ .
The local minimum is the point $(1.15; 3.77)$ .

Answer: a) $f^{'} (x) = 6 x^{2} - 5 - \frac{4}{x ^{2}}$ ; b) local minimum at $(1.15; 3.77)$

Exercise 35 MediumIB · P1 TZ0 November 2012

Consider the function $f (x) = 5 x^{3} - 4 x^{2} + x$ .

a) Calculate $f^{'} (x)$ .

b) Using your answer to part (a), find the $x$ -coordinate of

(i) the local maximum point;

(ii) the local minimum point.

Show answer

Answer: a) $f^{'} (x) = 15 x^{2} - 8 x + 1$ ; b) (i) local maximum at $x = \frac{1}{5} = 0.2$ ; (ii) local minimum at $x = \frac{1}{3} \approx 0.333$

Show working and answer

Differentiate term by term: $f^{'} (x) = 15 x^{2} - 8 x + 1.$

Set equal to zero and factorise: $15 x^{2} - 8 x + 1 = 0 \Rightarrow (5 x - 1) (3 x - 1) = 0.$
The stationary points are $x = \frac{1}{5}$ and $x = \frac{1}{3}$ .
We study the sign of $f^{'}$ : it is positive for $x < \frac{1}{5}$ , negative for $\frac{1}{5} < x < \frac{1}{3}$ and positive for $x > \frac{1}{3}$ .
(i) At $x = \frac{1}{5}$ it changes from increasing to decreasing: local maximum.
(ii) At $x = \frac{1}{3}$ it changes from decreasing to increasing: local minimum.

Answer: a) $f^{'} (x) = 15 x^{2} - 8 x + 1$ ; b) (i) local maximum at $x = \frac{1}{5} = 0.2$ ; (ii) local minimum at $x = \frac{1}{3} \approx 0.333$

Exercise 36 MediumIB · P1 TZ0 November 2007

A function is represented by the equation $f (x) = a x^{2} + \frac{4}{x} - 3$ .

a) Find $f^{'} (x)$ .

The function $f (x)$ has a local maximum at the point where $x = - 1$ .

b) Find the value of $a$ .

Show answer

Answer: a) $f^{'} (x) = 2 a x - \frac{4}{x ^{2}}$ ; b) $a = - 2$

Show working and answer

Differentiate term by term, writing $\frac{4}{x} = 4 x^{- 1}$ : $f^{'} (x) = 2 a x - 4 x^{- 2} = 2 a x - \frac{4}{x ^{2}} .$

At a local maximum the derivative is zero: $f^{'} (- 1) = 0$ .
Substitute $x = - 1$ : $2 a (- 1) - \frac{4}{( - 1 ) ^{2}} = - 2 a - 4 = 0 \Rightarrow a = - 2.$
Check: $f^{''} (x) = 2 a + \frac{8}{x ^{3}}$ , and with $a = - 2$ we get $f^{''} (- 1) = - 4 - 8 = - 12 < 0$ , so there is indeed a maximum at $x = - 1$ .

Answer: a) $f^{'} (x) = 2 a x - \frac{4}{x ^{2}}$ ; b) $a = - 2$

Exercise 37 MediumIB · P1 TZ2 May 2008

The function $f (x)$ is such that $f^{'} (x) < 0$ for $1 < x < 4$ .

At the point $P (4, 2)$ on the graph of $f (x)$ the gradient is zero.

a) Write down the equation of the tangent to the graph of $f (x)$ at $P$ .

b) State whether $f (4)$ is greater than, equal to, or less than $f (2)$ .

c) Given that $f (x)$ is increasing for $4 \leq x < 7$ , what can be said about the point $P$ ?

Show answer

Answer: a) $y = 2$ ; b) $f (4)$ is less than $f (2)$ ; c) $P$ is a local minimum

Show working and answer

Zero gradient means the tangent is horizontal.
It passes through $P (4, 2)$ , so its equation is $y = 2$ .

Since $f^{'} (x) < 0$ for $1 < x < 4$ , the function is decreasing on that interval.
Thus, for $2 < 4$ we have $f (2) > f (4)$ : $f (4)$ is less than $f (2)$ .

To the left of $x = 4$ the function decreases and to the right it increases, and at $P$ the gradient is zero.
Therefore $P$ is a local minimum of the graph.

Answer: a) $y = 2$ ; b) $f (4)$ is less than $f (2)$ ; c) $P$ is a local minimum

Exercise 38 MediumIB · P2 TZ0 November 2009

a) Sketch the graph of $y = 2^{x}$ for $- 2 \leq x \leq 3$ .

State clearly where the curve meets the $y$ -axis.

b) Write down the equation of the asymptote of the graph of $y = 2^{x}$ .

c) On the same axes, sketch the graph of $y = 3 + 2 x - x^{2}$ .

State clearly where the curve meets the $x$ -axis, and where it meets the $y$ -axis.

d) Using your GDC, solve the equation $3 + 2 x - x^{2} = 2^{x}$ .

e) Write down the maximum value of the function $f (x) = 3 + 2 x - x^{2}$ .

f) Use differential calculus to verify that your answer to part (e) is correct.

Show answer

Answer: a) graph of $y = 2^{x}$ with intercept $(0, 1)$ ; b) $y = 0$ ; c) graph of the parabola with intercepts $(- 1, 0)$ , $(3, 0)$ and $(0, 3)$ ; d) $x \approx - 0.857$ and $x \approx 1.77$ ; e) $4$ ; f) verified ( $x = 1$ , $f (1) = 4$ )

Show working and answer

Table: $2^{- 2} = 0.25$ , $2^{- 1} = 0.5$ , $2^{0} = 1$ , $2^{1} = 2$ , $2^{2} = 4$ , $2^{3} = 8$ .
The curve is increasing, always positive, and meets the $y$ -axis at $(0, 1)$ .

As $x \to - \infty$ , $2^{x} \to 0$ without ever reaching zero, so the horizontal asymptote is $y = 0$ (the $x$ -axis).

$y = 3 + 2 x - x^{2} = - (x - 3) (x + 1)$ is a parabola that opens downwards.
Intercepts with the $x$ -axis: $x = - 1$ and $x = 3$ . Intercept with the $y$ -axis: $(0, 3)$ . Vertex at $(1, 4)$ .

Axes with x from −2 to 3 and y from 0 to 8: curve y = 2^x passing through (0; 1) and approaching the x-axis from the left, and parabola y = 3 + 2x − x² with vertex (1; 4), meeting the y-axis at (0; 3) and the x-axis at (−1; 0) and (3; 0); the two curves intersect near x = −0.86 and x = 1.77

Using the GDC, find the points of intersection of $y = 3 + 2 x - x^{2}$ and $y = 2^{x}$ (the $x$ -coordinates are the solutions).
Solutions: $x \approx - 0.857$ and $x \approx 1.77$ .

The maximum of the parabola is its vertex: $f (1) = 3 + 2 - 1 = 4$ .
Maximum value: $4$ .

$f^{'} (x) = 2 - 2 x = 0 \Rightarrow x = 1$ .
$f^{''} (x) = - 2 < 0$ , so it is a maximum.
$f (1) = 3 + 2 - 1 = 4$ , which agrees with (e).

Exercise 39 MediumIB · P2 TZ1 May 2011

Consider the function $f (x) = x^{3} + \frac{48}{x}$ , $x \neq = 0$ .

a) Calculate $f (2)$ .

b) Draw the graph of the function $y = f (x)$ for $- 5 \leq x \leq 5$ and $- 200 \leq y \leq 200$ .

c) Find $f^{'} (x)$ .

d) Find $f^{'} (2)$ .

e) Write down the coordinates of the local maximum point on the graph of $f$ .

f) Find the range of $f$ .

g) Find the gradient of the tangent to the graph of $f$ at $x = 1$ .

There is a second point on the graph of $f$ at which the tangent is parallel to the tangent at $x = 1$ .

h) Find the $x$ -coordinate of this point.

Show answer

Answer: a) $32$ ; b) graph with asymptote $x = 0$ ; c) $f^{'} (x) = 3 x^{2} - \frac{48}{x ^{2}}$ ; d) $0$ ; e) $(- 2, - 32)$ ; f) $f (x) \leq - 32$ or $f (x) \geq 32$ ; g) $- 45$ ; h) $x = - 1$

Show working and answer

a) $f (2) = 2^{3} + \frac{48}{2} = 8 + 24 = 32.$

Using the GDC, draw $y = x^{3} + \frac{48}{x}$ in the required window. There are two branches, separated by the vertical asymptote $x = 0$ .
Left branch: rises from below to a local maximum at $(- 2, - 32)$ , and then falls towards $- \infty$ as it approaches $x = 0$ . Right branch: falls from $+ \infty$ to a local minimum at $(2, 32)$ , and then rises.

Graph of f(x) = x³ + 48/x in the window −5 ≤ x ≤ 5, −200 ≤ y ≤ 200: left branch with a local maximum at (−2; −32) and vertical asymptote at x = 0; right branch with a local minimum at (2; 32), rising to about (5; 135)

c) $f^{'} (x) = 3 x^{2} - \frac{48}{x ^{2}} .$

d) $f^{'} (2) = 3 \cdot 4 - \frac{48}{4} = 12 - 12 = 0.$

$f^{'} (x) = 0 \Rightarrow 3 x^{4} = 48 \Rightarrow x^{4} = 16 \Rightarrow x = \pm 2$ .
At $x = 2$ there is a local minimum ( $f (2) = 32$ ), so the local maximum is at $x = - 2$ : $f (- 2) = - 8 - 24 = - 32.$
Local maximum point: $(- 2, - 32)$ .

Left branch ( $x < 0$ ): reaches $- 32$ as a maximum and tends to $- \infty$ : values in $] - \infty, - 32]$ .
Right branch ( $x > 0$ ): minimum $32$ and increases to $+ \infty$ : values in $[32, + \infty [$ .
Range: $f (x) \leq - 32$ or $f (x) \geq 32$ .

g) $f^{'} (1) = 3 - 48 = - 45.$

We look for another $x$ with $f^{'} (x) = - 45$ : $3 x^{2} - \frac{48}{x ^{2}} = - 45 \Rightarrow 3 x^{4} + 45 x^{2} - 48 = 0 \Rightarrow x^{4} + 15 x^{2} - 16 = 0.$
With $u = x^{2}$ : $u^{2} + 15 u - 16 = (u - 1) (u + 16) = 0$ , so $u = 1$ (the other root is negative) and $x = \pm 1$ .
The point other than $x = 1$ has $x = - 1$ (check: $f^{'} (- 1) = 3 - 48 = - 45$ ).

Answer: a) $32$ ; b) graph with asymptote $x = 0$ ; c) $f^{'} (x) = 3 x^{2} - \frac{48}{x ^{2}}$ ; d) $0$ ; e) $(- 2, - 32)$ ; f) $f (x) \leq - 32$ or $f (x) \geq 32$ ; g) $- 45$ ; h) $x = - 1$

Exercise 40 HardIB · P2 TZ1 May 2019

The function $f (x) = \frac{1}{3} x^{3} + \frac{1}{2} x^{2} + k x + 5$ has a local maximum and a local minimum.

The local maximum is at $x = - 3$ .

a) Show that $k = - 6$ .

b) Find the coordinates of the local minimum.

c) Write down the interval where the gradient of the graph of $f (x)$ is negative.

d) Determine the equation of the normal at $x = - 2$ in the form $y = m x + c$ .

Show answer

Answer: a) $k = - 6$ (shown); b) $(2, - \frac{7}{3})$ ; c) $] - 3, 2 [$ ; d) $y = \frac{1}{4} x + \frac{101}{6}$

Show working and answer

Differentiate: $f^{'} (x) = x^{2} + x + k$ .
At a local maximum the derivative is zero. With $x = - 3$ : $f^{'} (- 3) = 9 - 3 + k = 0 \Rightarrow k = - 6.$

With $k = - 6$ : $f^{'} (x) = x^{2} + x - 6 = (x + 3) (x - 2)$ , which is zero at $x = - 3$ and $x = 2$ .
The maximum is at $x = - 3$ , so the local minimum is at $x = 2$ (there $f^{'}$ changes from negative to positive).
We calculate its $y$ -coordinate: $f (2) = \frac{8}{3} + 2 - 12 + 5 = - \frac{7}{3} \approx - 2.33.$

The gradient is $f^{'} (x) = (x + 3) (x - 2)$ , which is negative between its two roots: $- 3 < x < 2$ .

$f (- 2) = - \frac{8}{3} + 2 + 12 + 5 = \frac{49}{3}$ and $f^{'} (- 2) = 4 - 2 - 6 = - 4$ .
The gradient of the normal is $- \frac{1}{- 4} = \frac{1}{4}$ .
Equation of the normal: $y - \frac{49}{3} = \frac{1}{4} (x + 2) \Rightarrow y = \frac{1}{4} x + \frac{101}{6} .$

Answer: a) $k = - 6$ (shown); b) $(2, - \frac{7}{3})$ ; c) $] - 3, 2 [$ ; d) $y = \frac{1}{4} x + \frac{101}{6}$

Exercise 41 HardIB · P2 TZ2 May 2019

Consider the function $f (x) = \frac{1}{3} x^{3} + \frac{3}{4} x^{2} - x - 1$ .

a) Find $f (2)$ .

b) Write down the $y$ -intercept of the graph of $y = f (x)$ .

c) Sketch the graph of $y = f (x)$ for $- 3 \leq x \leq 3$ and $- 4 \leq y \leq 12$ .

d) Find $f^{'} (x)$ .

e) Find the gradient of the graph of $y = f (x)$ at $x = 2$ .

f) Find the equation of the tangent to the graph of $y = f (x)$ at $x = 2$ . Give the equation in the form $a x + b y + d = 0$ where $a$ , $b$ and $d \in Z$ .

The function has a local maximum at $x = p$ and a local minimum at $x = q$ .

g) Use the derivative of $f (x)$ to find the values of $p$ and $q$ .

h) Determine the range of $f (x)$ for $p \leq x \leq q$ .

Show answer

Answer: a) $\frac{8}{3} \approx 2.67$ ; b) $(0, - 1)$ ; c) graph with local maximum at $x = - 2$ and local minimum at $x = 0.5$ ; d) $f^{'} (x) = x^{2} + \frac{3}{2} x - 1$ ; e) $6$ ; f) $18 x - 3 y - 28 = 0$ ; g) $p = - 2$ , $q = 0.5$ ; h) $- 1.27 \leq f (x) \leq 1.33$

Show working and answer

$f (2) = \frac{8}{3} + \frac{3}{4} \cdot 4 - 2 - 1 = \frac{8}{3} + 3 - 2 - 1 = \frac{8}{3} \approx 2.67$ .

With $x = 0$ : $f (0) = - 1$ . The $y$ -intercept is $(0, - 1)$ .

Using the GDC, draw the graph in the window $- 3 \leq x \leq 3$ , $- 4 \leq y \leq 12$ . It is a cubic with a local maximum at $x = - 2$ , a local minimum at $x = 0.5$ , and it ends at $x = 3$ near $y = 11.75$ .

Graph of f(x) = x³/3 + 3x²/4 − x − 1 for −3 ≤ x ≤ 3 and −4 ≤ y ≤ 12 on a grid, with a local maximum near x = −2 and a local minimum near x = 0.5

$f^{'} (x) = x^{2} + \frac{3}{2} x - 1$ .

$f^{'} (2) = 4 + 3 - 1 = 6$ .

The tangent has gradient $6$ and passes through $(2, \frac{8}{3})$ : $y - \frac{8}{3} = 6 (x - 2) \Rightarrow y = 6 x - \frac{28}{3} .$
Multiply by $3$ and move all terms to one side: $18 x - 3 y - 28 = 0.$

We solve $f^{'} (x) = 0$ : $x^{2} + \frac{3}{2} x - 1 = 0 \Rightarrow 2 x^{2} + 3 x - 2 = 0 \Rightarrow x = \frac{- 3 \pm 5}{4},$ that is, $x = - 2$ or $x = \frac{1}{2}$ .
$f^{'}$ changes from positive to negative at $x = - 2$ (maximum) and from negative to positive at $x = \frac{1}{2}$ (minimum). So $p = - 2$ and $q = 0.5$ .

$f (- 2) = - \frac{8}{3} + 3 + 2 - 1 = \frac{4}{3} \approx 1.33$ .
$f (\frac{1}{2}) = \frac{1}{24} + \frac{3}{16} - \frac{1}{2} - 1 = - \frac{61}{48} \approx - 1.27$ .
Between $p$ and $q$ the function is decreasing, so the range runs from the value at $q$ to the value at $p$ : $- \frac{61}{48} \leq f (x) \leq \frac{4}{3} .$

Exercise 42 HardIB · P2 TZ1 May 2018

Consider the function $f (x) = \frac{48}{x} + k x^{2} - 58$ , where $x > 0$ and $k$ is a constant.

The graph of the function passes through the point with coordinates $(4, 2)$ .

a) Find the value of $k$ .

b) Using your value of $k$ , find $f^{'} (x)$ .

$P$ is the minimum point of the graph of $f (x)$ .

c) Use your answer to part (b) to show that the minimum value of $f (x)$ is $- 22$ .

d) Write down the two values of $x$ that satisfy $f (x) = 0$ .

e) Sketch the graph of $y = f (x)$ for $0 < x < 6$ and $- 30 \leq y \leq 60$ . Clearly indicate the minimum point $P$ and the $x$ -intercepts on your graph.

Show answer

Answer: a) $k = 3$ ; b) $f^{'} (x) = - \frac{48}{x ^{2}} + 6 x$ ; c) minimum $f (2) = - 22$ (shown); d) $x \approx 0.861$ and $x \approx 3.90$ ; e) graph with minimum $P (2, - 22)$ and intercepts at $x \approx 0.861$ and $x \approx 3.90$

Show working and answer

The graph passes through $(4, 2)$ , so $f (4) = 2$ : $\frac{48}{4} + 16 k - 58 = 2 \Rightarrow 16 k = 48 \Rightarrow k = 3.$

With $k = 3$ : $f (x) = 48 x^{- 1} + 3 x^{2} - 58$ .
Differentiate: $f^{'} (x) = - \frac{48}{x ^{2}} + 6 x .$

The minimum satisfies $f^{'} (x) = 0$ : $6 x = \frac{48}{x ^{2}} \Rightarrow x^{3} = 8 \Rightarrow x = 2.$
$f^{'}$ is negative for $x < 2$ and positive for $x > 2$ (at $x = 1$ it equals $- 42$ and at $x = 3$ it equals $\approx 12.7$ ), so it is a minimum.
Minimum value: $f (2) = \frac{48}{2} + 3 \cdot 4 - 58 = 24 + 12 - 58 = - 22.$

Using the GDC, the solutions of $f (x) = 0$ are $x \approx 0.861$ and $x \approx 3.90$ .

The graph is drawn using the GDC in the window $0 < x < 6$ , $- 30 \leq y \leq 60$ : it falls from the top (asymptote at $x = 0$ ), cuts the $x$ -axis at $x \approx 0.861$ , reaches the minimum $P (2, - 22)$ , cuts the axis again at $x \approx 3.90$ and rises to near $y = 58$ at $x = 6$ .

Graph of f(x) = 48/x + 3x² − 58 for 0 < x < 6 and −30 ≤ y ≤ 60 on a grid; cuts the x-axis at (0.861; 0) and (3.90; 0) and has a minimum P at (2, −22)

Exercise 43 HardIB · P2 TZ2 May 2009

A function is defined by $f (x) = \frac{5}{x ^{2}} + 3 x + c$ , $x \neq = 0$ , $c \in Z$ .

a) Write down an expression for $f^{'} (x)$ .

The graph of $f$ passes through the point $P (1, 4)$ .

b) Find the value of $c$ .

c) There is a local minimum at the point $Q$ .

(i) Find the coordinates of $Q$ .

(ii) Find the set of values of $x$ for which the function is decreasing.

Let $T$ be the tangent to the graph of $f$ at $P$ .

d) (i) Show that the gradient of $T$ is $- 7$ .

(ii) Find the equation of $T$ .

e) $T$ meets the graph again at $R$ . Use your graphic display calculator to find the coordinates of $R$ .

Show answer

Answer: a) $f^{'} (x) = - \frac{10}{x ^{3}} + 3$ ; b) $c = - 4$ ; c) (i) $Q (1.49, 2.72)$ ; (ii) $0 < x < 1.49$ ; d) (i) gradient $- 7$ (shown); (ii) $y = - 7 x + 11$ ; e) $R (- 0.5, 14.5)$

Show working and answer

Write $f (x) = 5 x^{- 2} + 3 x + c$ and differentiate: $f^{'} (x) = - \frac{10}{x ^{3}} + 3.$

The graph passes through $P (1, 4)$ : $f (1) = 5 + 3 + c = 4$ , so $c = - 4$ .

(i) At the local minimum $f^{'} (x) = 0$ : $- \frac{10}{x ^{3}} + 3 = 0 \Rightarrow x^{3} = \frac{10}{3} \Rightarrow x \approx 1.49.$
$f (1.49) \approx 2.72$ , so $Q (1.49, 2.72)$ .
(ii) The function is decreasing where $f^{'} (x) < 0$ , that is, $\frac{10}{x ^{3}} > 3$ . This occurs for $0 < x < 1.49$ (for $x < 0$ the derivative is positive).

(i) $f^{'} (1) = - 10 + 3 = - 7$ , which is the gradient of $T$ .
(ii) $T$ passes through $P (1, 4)$ with gradient $- 7$ : $y - 4 = - 7 (x - 1) \Rightarrow y = - 7 x + 11.$

Using the GDC, the intersections of $f (x) = \frac{5}{x ^{2}} + 3 x - 4$ with $y = - 7 x + 11$ are $x = 1$ (the point $P$ ) and $x = - 0.5$ .
Then $y = - 7 (- 0.5) + 11 = 14.5$ and the point is $R (- 0.5, 14.5)$ .

Answer: a) $f^{'} (x) = - \frac{10}{x ^{3}} + 3$ ; b) $c = - 4$ ; c) (i) $Q (1.49, 2.72)$ ; (ii) $0 < x < 1.49$ ; d) (i) gradient $- 7$ (shown); (ii) $y = - 7 x + 11$ ; e) $R (- 0.5, 14.5)$

Exercise 44 HardIB · P2 TZ2 May 2017

Consider the function $f (x) = - x^{4} + a x^{2} + 5$ , where $a$ is a constant.

Part of the graph of $y = f (x)$ is shown below.

Hand-drawn graph of y = −x⁴ + 4x² + 5 in the shape of an inverted M, with no grid or numerical marks

a) Write down the point where the graph cuts the $y$ -axis.

b) Find $f^{'} (x)$ .

It is known that at the point where $x = 2$ the tangent to the graph of $y = f (x)$ is horizontal.

c) (i) Show that $a = 8$ .

(ii) Find $f (2)$ .

The graph of $y = f (x)$ has two other points at which the tangent is horizontal.

d) Write down

(i) the $x$ -coordinate of these two points;

(ii) the intervals on which the gradient of the graph of $y = f (x)$ is positive.

e) Write down the range of $f (x)$ .

f) Write down the number of possible solutions of the equation $f (x) = 5$ .

g) The equation $f (x) = m$ , where $m \in R$ , has four solutions.

Find the possible values of $m$ .

Show answer

Answer: a) $(0, 5)$ ; b) $f^{'} (x) = - 4 x^{3} + 2 a x$ ; c) (i) $a = 8$ (shown); (ii) $f (2) = 21$ ; d) (i) $x = 0$ and $x = - 2$ ; (ii) $] - \infty, - 2 [\cup] 0, 2 [$ ; e) $f (x) \leq 21$ , that is, $] - \infty, 21]$ ; f) $3$ solutions; g) $5 < m < 21$

Show working and answer

With $x = 0$ : $f (0) = 5$ . The $y$ -intercept is $(0, 5)$ .

$f^{'} (x) = - 4 x^{3} + 2 a x$ .

(i) A horizontal tangent at $x = 2$ means $f^{'} (2) = 0$ : $- 4 \cdot 8 + 4 a = 0 \Rightarrow - 32 + 4 a = 0 \Rightarrow a = 8.$
(ii) $f (2) = - 16 + 8 \cdot 4 + 5 = 21$ .

(i) With $a = 8$ : $f^{'} (x) = - 4 x^{3} + 16 x = - 4 x (x - 2) (x + 2)$ , which is zero at $x = - 2$ , $x = 0$ and $x = 2$ . The other two points have $x = 0$ and $x = - 2$ .
(ii) The gradient is positive ( $f^{'} (x) > 0$ ) when $x < - 2$ or $0 < x < 2$ (for example, $f^{'} (- 3) = 60 > 0$ and $f^{'} (1) = 12 > 0$ ): $] - \infty, - 2 [\cup] 0, 2 [.$

The local maxima are $f (\pm 2) = 21$ and the local minimum is $f (0) = 5$ ; moreover $f (x) \to - \infty$ as $x \to \pm \infty$ .
The greatest value the function attains is $21$ , so the range is $f (x) \leq 21$ .

$- x^{4} + 8 x^{2} + 5 = 5 \Rightarrow x^{2} (8 - x^{2}) = 0$ , so $x = 0$ , $x = 8$ or $x = - 8$ .
There are $3$ solutions.

The horizontal line $y = m$ cuts the graph four times when it lies between the local minimum ( $5$ ) and the local maxima ( $21$ ).
Therefore $5 < m < 21$ .

Exercise 45 HardIB · P2 TZ0 November 2017

A function $f$ is given by $f (x) = (2 x + 2) (5 - x^{2})$ .

a) Find the exact value of each of the zeros of $f$ .

b) (i) Expand the expression for $f (x)$ .

(ii) Find $f^{'} (x)$ .

c) Use your answer to part (b)(ii) to find the values of $x$ for which $f$ is increasing.

d) Draw an accurate graph of $f$ for $- 3 \leq x \leq 3$ and $- 40 \leq y \leq 20$ . Use a scale of 2 cm to represent 1 unit on the $x$ -axis and 1 cm to represent 5 units on the $y$ -axis.

The graph of the function $g (x) = 5^{x} + 6 x - 6$ intersects the graph of $f$ .

e) Write down the coordinates of the point of intersection.

Show answer

Answer: a) $x = - 1$ , $x = 5$ and $x = - 5$ ; b) (i) $f (x) = - 2 x^{3} - 2 x^{2} + 10 x + 10$ ; (ii) $f^{'} (x) = - 6 x^{2} - 4 x + 10$ ; c) $- \frac{5}{3} < x < 1$ ; d) graph with zeros $- 5, - 1, 5$ , $y$ -intercept $(0, 10)$ , local minimum at $x = - \frac{5}{3}$ and local maximum $(1, 16)$ ; e) $(1.49, 13.9)$

Show working and answer

A product is zero if one of its factors is zero: $2 x + 2 = 0 or 5 - x^{2} = 0.$
From the first, $x = - 1$ ; from the second, $x = \pm 5$ .

(i) Multiply the factors: $f (x) = 10 x - 2 x^{3} + 10 - 2 x^{2} = - 2 x^{3} - 2 x^{2} + 10 x + 10.$
(ii) Differentiate term by term: $f^{'} (x) = - 6 x^{2} - 4 x + 10.$

$f$ is increasing when $f^{'} (x) > 0$ : $- 6 x^{2} - 4 x + 10 > 0 ⟺ 3 x^{2} + 2 x - 5 < 0 ⟺ (3 x + 5) (x - 1) < 0.$
The roots of $f^{'}$ are $x = - \frac{5}{3}$ and $x = 1$ , and the parabola $3 x^{2} + 2 x - 5$ is negative between them.
Therefore, $f$ is increasing for $- \frac{5}{3} < x < 1$ .

Draw the graph on the interval $- 3 \leq x \leq 3$ (2 cm per unit in $x$ , 1 cm per 5 units in $y$ ), marking the zeros $x = - 5 \approx - 2.24$ , $x = - 1$ and $x = 5 \approx 2.24$ , and the $y$ -intercept, $f (0) = 10$ .
The curve starts at $(- 3, 16)$ , falls, cutting the $x$ -axis at $- 2.24$ , reaches a local minimum at $x = - \frac{5}{3}$ (height approximately $- 3.0$ ), rises, cutting the $x$ -axis at $- 1$ and the $y$ -axis at $10$ , reaches a local maximum at $(1, 16)$ and falls, cutting the $x$ -axis at $2.24$ , down to $f (3) = - 32$ .

Graph of f(x) = (2x + 2)(5 − x²) on a grid, for −3 ≤ x ≤ 3 and −40 ≤ y ≤ 20; the three zeros and the y-intercept are marked

Using the graphic display calculator, we look for the intersection of $y = 5^{x} + 6 x - 6$ with $y = f (x)$ , that is, we solve $5^{x} + 6 x - 6 = (2 x + 2) (5 - x^{2}) .$
The $x$ -coordinate is $x \approx 1.487$ and the $y$ -coordinate is $y = f (1.487) \approx 13.87$ .
The point of intersection is approximately $(1.49, 13.9)$ .

Exercise 46 HardIB · P2 TZ1 May 2015

Consider the function $f (x) = \frac{96}{x ^{2}} + k x$ , where $k$ is a constant and $x \neq = 0$ .

a) Write $f^{'} (x)$ .

The graph of $y = f (x)$ has a local minimum at $x = 4$ .

b) Show that $k = 3$ .

c) Calculate $f (2)$ .

d) Calculate $f^{'} (2)$ .

e) Find the equation of the normal to the graph of $y = f (x)$ at the point where $x = 2$ . Give your answer in the form $a x + b y + d = 0$ where $a, b, d \in Z$ .

f) Draw the graph of $y = f (x)$ for $- 5 \leq x \leq 10$ and $- 10 \leq y \leq 100$ .

g) Write down the coordinates of the point where the graph of $y = f (x)$ cuts the $x$ -axis.

h) State the values of $x$ for which $f (x)$ is decreasing.

Show answer

Answer: a) $f^{'} (x) = - \frac{192}{x ^{3}} + k$ ; b) $k = 3$ ; c) $f (2) = 30$ ; d) $f^{'} (2) = - 21$ ; e) $x - 21 y + 628 = 0$ ; f) graph with vertical asymptote $x = 0$ , x-intercept $(- 3.17; 0)$ and local minimum $(4, 18)$ ; g) $(- 3.17; 0)$ ; h) $f$ decreasing for $0 < x < 4$

Show working and answer

Write $f (x) = 96 x^{- 2} + k x$ and differentiate: $f^{'} (x) = - 192 x^{- 3} + k = - \frac{192}{x ^{3}} + k .$

At a local minimum the derivative is zero, so $f^{'} (4) = 0$ : $- \frac{192}{64} + k = 0 \Rightarrow - 3 + k = 0.$
Hence $k = 3$ .

With $k = 3$ : $f (2) = \frac{96}{4} + 3 \cdot 2 = 24 + 6 = 30.$

With $k = 3$ : $f^{'} (2) = - \frac{192}{8} + 3 = - 24 + 3 = - 21.$

The point is $(2, 30)$ . The gradient of the tangent is $- 21$ , so the gradient of the normal is $\frac{1}{21}$ .
The equation of the normal is $y - 30 = \frac{1}{21} (x - 2) \Rightarrow 21 y - 630 = x - 2 \Rightarrow x - 21 y + 628 = 0.$

Draw $f (x) = \frac{96}{x ^{2}} + 3 x$ for $- 5 \leq x \leq 10$ and $- 10 \leq y \leq 100$ . The $y$ -axis ( $x = 0$ ) is a vertical asymptote.
For $x < 0$ , the branch cuts the $x$ -axis at $(- 3.17; 0)$ and rises towards $+ \infty$ as it approaches $x = 0$ from the left.
For $x > 0$ , the branch falls from $+ \infty$ to the local minimum $(4, 18)$ and then rises gently to $f (10) = 31.56$ .

Graph of y = 96/x² + 3x for −5 ≤ x ≤ 10, with vertical asymptote x = 0, the x-intercept and the local minimum (4; 18)

The graph cuts the $x$ -axis where $f (x) = 0$ : $\frac{96}{x ^{2}} + 3 x = 0 \Rightarrow 96 + 3 x^{3} = 0 \Rightarrow x^{3} = - 32.$
Hence $x = - 332 \approx - 3.17$ and the point is $(- 3.17; 0)$ .

For $f$ to be decreasing, $f^{'} (x) < 0$ : $- \frac{192}{x ^{3}} + 3 < 0 ⟺ \frac{192}{x ^{3}} > 3.$
If $x < 0$ , $\frac{192}{x ^{3}}$ is negative, so the inequality does not hold (in that region $f^{'} > 0$ ).
If $x > 0$ , the inequality is equivalent to $x^{3} < 64$ , that is, $x < 4$ .
Therefore $f$ is decreasing for $0 < x < 4$ .

Exercise 47 HardIB · P2 TZ0 November 2015

The following diagram shows the graph of the function $f (x) = n x^{3} + p x^{2} + q x + r$ , $n \neq = 0$ , for $- 5 \leq x \leq 7$ .

Graph of f(x) = 10x³ − 30x² − 240x + 150 for −5 ≤ x ≤ 7, with local maximum (−2; 430), local minimum (4; −650) and the point P where it cuts the x-axis

a) State whether the function is increasing or decreasing at $x = - 3$ . Give a reason for your answer.

b) Write down the value of $r$ .

The values of $p$ and $q$ are such that $f (x) = n x^{3} - 30 x^{2} - 240 x + r$ .

c) Find $f^{'} (x)$ .

d) Write down the coordinates of the local minimum.

e) Show that the value of $n$ is 10.

f) (i) Calculate $f^{'} (- 1)$ .

(ii) Find the equation of the tangent to the graph at the point $(- 1, 350)$ . Give your answer in the form $a x + b y + d = 0$ .

(iii) State the gradient of the normal to the graph at $x = - 1$ .

The graph of the function cuts the $x$ -axis at the point $P$ , as shown in the figure.

g) Use the GDC to find the $x$ -coordinate of $P$ .

h) Let $g (x) = 100 x + 400$ , for $- 5 \leq x \leq 7$ . Use the GDC to find the values of $x$ for which $f (x) = g (x)$ , within the given domain.

Show answer

Answer: a) increasing at $x = - 3$ ; b) $r = 150$ ; c) $f^{'} (x) = 3 n x^{2} - 60 x - 240$ ; d) $(4, - 650)$ ; e) $n = 10$ ; f) (i) $f^{'} (- 1) = - 150$ ; (ii) $150 x + y - 200 = 0$ ; (iii) $\frac{1}{150}$ ; g) $x \approx 6.39$ ; h) $x \approx - 3.97$ and $x \approx - 0.809$

Show working and answer

At $x = - 3$ the graph is to the left of the local maximum $x = - 2$ , and in that region the curve rises.
Hence $f$ is increasing at $x = - 3$ .

$f (0) = r$ , and the graph cuts the $y$ -axis at $(0, 150)$ .
Hence $r = 150$ .

Differentiate $f (x) = n x^{3} - 30 x^{2} - 240 x + r$ : $f^{'} (x) = 3 n x^{2} - 60 x - 240.$

From the graph, the local minimum is the point $(4, - 650)$ .

At the local minimum, $f^{'} (4) = 0$ : $3 n \cdot 16 - 60 \cdot 4 - 240 = 0 \Rightarrow 48 n - 480 = 0.$
Hence $n = 10$ .

(i) With $n = 10$ , $f^{'} (x) = 30 x^{2} - 60 x - 240$ , and $f^{'} (- 1) = 30 + 60 - 240 = - 150.$
(ii) Check the point: $f (- 1) = - 10 - 30 + 240 + 150 = 350$ . The tangent has gradient $- 150$ and passes through $(- 1, 350)$ : $y - 350 = - 150 (x + 1) \Rightarrow 150 x + y - 200 = 0.$
(iii) The gradient of the normal is the negative reciprocal of $- 150$ : $\frac{1}{150} .$

Using the GDC, look for the intersection of $f (x) = 10 x^{3} - 30 x^{2} - 240 x + 150$ with the $x$ -axis near $x = 6$ .
We obtain $x \approx 6.39$ .

Equate the two functions: $10 x^{3} - 30 x^{2} - 240 x + 150 = 100 x + 400.$
Using the GDC, draw both graphs and find their intersections within $- 5 \leq x \leq 7$ .
The values are $x \approx - 3.97$ and $x \approx - 0.809$ . (There is a third solution, $x \approx 7.78$ , which lies outside the domain.)

Exercise 48 HardIB · P2 TZ1 May 2012

Consider the function $f (x) = - \frac{1}{3} x^{3} + \frac{5}{3} x^{2} - x - 3$ .

a) Draw the graph of $y = f (x)$ for $- 3 \leq x \leq 6$ and $- 10 \leq y \leq 10$ , clearly showing the intercepts with the axes and the local maximum and minimum points. Use a scale of 2 cm to represent 1 unit on the $x$ -axis, and a scale of 1 cm to represent 1 unit on the $y$ -axis.

b) Find the value of $f (- 1)$ .

c) Write down the coordinates of the $y$ -intercept of the graph of $f (x)$ .

d) Calculate $f^{'} (x)$ .

e) Show that $f^{'} (- 1) = - \frac{16}{3}$ .

f) Explain what $f^{'} (- 1)$ represents.

g) Find the equation of the tangent to the graph of $f (x)$ at the point where $x = - 1$ .

h) Draw the tangent to the graph of $f (x)$ at $x = - 1$ on your diagram from part (a).

$P$ and $Q$ are points on the curve such that the tangents to the curve at these points are horizontal. The $x$ -coordinate of $P$ is $a$ , and the $x$ -coordinate of $Q$ is $b$ , where $b > a$ .

(i) Write down the value of

(i) $a$ ;

(ii) $b$ .

j) Describe the behaviour of $f (x)$ for $a < x < b$ .

Show answer

Answer: a) graph with intercepts $(- 1, 0)$ , $(3, 0)$ , $(0, - 3)$ , minimum $(0.333; - 3.16)$ and maximum $(3, 0)$ ; b) $0$ ; c) $(0, - 3)$ ; d) $f^{'} (x) = - x^{2} + \frac{10}{3} x - 1$ ; e) shown; f) gradient of the tangent at $x = - 1$ ; g) $y = - \frac{16}{3} x - \frac{16}{3}$ ; h) tangent drawn; i) (i) $a = \frac{1}{3}$ ; (ii) $b = 3$ ; j) $f$ is increasing

Show working and answer

Using the GDC, the graph of $f$ is obtained in the required window. Intercepts with the axes: $(- 1, 0)$ , $(3, 0)$ and $(0, - 3)$ .
Extrema: local minimum at $(0.333; - 3.16)$ and local maximum at $(3, 0)$ .
Scale: $2$ cm per unit on the $x$ -axis and $1$ cm per unit on the $y$ -axis. The graph crosses the $x$ -axis at $x = - 1$ , touches the $x$ -axis at $x = 3$ (maximum) and leaves the window downwards to the right.

Graph of f(x) = −(1/3)x³ + (5/3)x² − x − 3 on −3 ≤ x ≤ 6, −10 ≤ y ≤ 10, with a scale of 2 units of x per 1 unit of y: crosses the x-axis at −1, touches the x-axis at (3; 0) as a local maximum, crosses the y-axis at (0; −3), has a local minimum at (0.33; −3.16), and the tangent at x = −1, with gradient −16/3, passes through (−1; 0)

b) $f (- 1) = - \frac{1}{3} (- 1)^{3} + \frac{5}{3} (- 1)^{2} - (- 1) - 3 = \frac{1}{3} + \frac{5}{3} + 1 - 3 = 0.$

$f (0) = - 3$ , so the $y$ -intercept is $(0, - 3)$ .

d) $f^{'} (x) = - x^{2} + \frac{10}{3} x - 1.$

e) $f^{'} (- 1) = - (- 1)^{2} + \frac{10}{3} (- 1) - 1 = - 1 - \frac{10}{3} - 1 = - \frac{16}{3} .$

$f^{'} (- 1)$ is the gradient of the tangent to the graph at the point with $x = - 1$ , that is, the instantaneous rate of change of $f$ at that point.
Since it is negative, $f$ is decreasing at $x = - 1$ .

The point of tangency is $(- 1, 0)$ and the gradient is $- \frac{16}{3}$ : $y - 0 = - \frac{16}{3} (x + 1) \Rightarrow y = - \frac{16}{3} x - \frac{16}{3} .$

Draw the line $y = - \frac{16}{3} x - \frac{16}{3}$ on the diagram from (a): it passes through $(- 1, 0)$ and $(0, - 5.33)$ , and slopes downwards to the right, touching the curve only at $(- 1, 0)$ .

Horizontal tangent: $f^{'} (x) = 0$ : $- x^{2} + \frac{10}{3} x - 1 = 0 \Rightarrow 3 x^{2} - 10 x + 3 = 0 \Rightarrow x = \frac{10 \pm 8}{6} .$
(i) $a = \frac{1}{3}$ ; (ii) $b = 3$ .

$f^{'} (x) = - x^{2} + \frac{10}{3} x - 1$ is a parabola that opens downwards, positive between its roots $\frac{1}{3}$ and $3$ . (For example, $f^{'} (1) = \frac{4}{3} > 0$ .)
Therefore, for $\frac{1}{3} < x < 3$ the function $f$ is increasing.

Exercise 49 HardIB · P2 TZ2 May 2011

The function $f (x)$ is defined by $f (x) = 1.5 x + 4 + \frac{6}{x}$ , $x \neq = 0$ .

a) Write down the equation of the vertical asymptote.

b) Calculate $f^{'} (x)$ .

c) Find the gradient of the graph of the function at $x = - 1$ .

d) Using your answer to part (c), decide whether the function $f (x)$ is increasing or decreasing at $x = - 1$ . Justify your answer.

e) Draw the graph of $f (x)$ for $- 10 \leq x \leq 10$ and $- 20 \leq y \leq 20$ .

$P_{1}$ is the local maximum point and $P_{2}$ is the local minimum point on the graph of $f (x)$ .

f) Using your GDC, write down the coordinates of

(i) $P_{1}$ ;

(ii) $P_{2}$ .

g) Using your graph from (e), determine the range of the function $f (x)$ for $- 10 \leq x \leq 10$ .

Show answer

Answer: a) $x = 0$ ; b) $f^{'} (x) = 1.5 - \frac{6}{x ^{2}}$ ; c) $- 4.5$ ; d) decreasing (negative gradient); e) graph with asymptote $x = 0$ ; f) (i) $P_{1} (- 2, - 2)$ ; (ii) $P_{2} (2, 10)$ ; g) $y \leq - 2$ or $y \geq 10$

Show working and answer

$\frac{6}{x}$ is not defined at $x = 0$ and grows without bound near it.
Vertical asymptote: $x = 0$ .

Write $f (x) = 1.5 x + 4 + 6 x^{- 1}$ and differentiate: $f^{'} (x) = 1.5 - \frac{6}{x ^{2}} .$

c) $f^{'} (- 1) = 1.5 - \frac{6}{1} = - 4.5.$

The gradient at $x = - 1$ is $- 4.5 < 0$ .
Therefore $f$ is decreasing at $x = - 1$ .

Using the GDC, draw $f$ in the required window. There are two branches, separated by the asymptote $x = 0$ : the left one rises to a local maximum at $(- 2, - 2)$ and then falls to $- \infty$ ; the right one falls from $+ \infty$ to a local minimum at $(2, 10)$ and then rises.

Graph of f(x) = 1.5x + 4 + 6/x for −10 ≤ x ≤ 10 and −20 ≤ y ≤ 20, with grid and the vertical asymptote x = 0 dashed: local maximum P1(−2; −2) on the left branch and local minimum P2(2; 10) on the right branch

Using the GDC (equivalent to $f^{'} (x) = 0 \Rightarrow x^{2} = 4$ ) we obtain $x = \pm 2$ .
(i) $P_{1} = (- 2, - 2)$ (local maximum), since $f (- 2) = - 3 + 4 - 3 = - 2$ .
(ii) $P_{2} = (2, 10)$ (local minimum), since $f (2) = 3 + 4 + 3 = 10$ .

For $- 10 \leq x < 0$ : $f$ rises from $f (- 10) = - 11.6$ to $- 2$ and then falls to $- \infty$ ; the values are $y \leq - 2$ .
For $0 < x \leq 10$ : $f$ falls from $+ \infty$ to $10$ and then rises to $f (10) = 19.6$ ; the values are $y \geq 10$ .
Range: $y \leq - 2$ or $y \geq 10$ .

Exercise 50 HardIB · P2 TZ1 May 2010

The diagram shows part of the graph of $f (x) = 4 x^{3} - 9 x^{2} - 12 x + 3$ .

Part of the graph of a cubic f(x) = 4x³ − 9x² − 12x + 3, with the x- and y-axes drawn as arrows and no grid or numerical marks: it rises, crossing the x-axis to the left of the y-axis; it has a local maximum to the left of the y-axis; it crosses the y-axis above the origin; it falls to a local minimum below the x-axis and rises again, crossing the x-axis to the right

a) Write down the values of $x$ where the graph of $f (x)$ crosses the $x$ -axis.

b) Write down $f^{'} (x)$ .

c) Find the value of the local maximum of $y = f (x)$ .

Let $P$ be the point where the graph of $f (x)$ crosses the $y$ -axis.

d) Write down the coordinates of $P$ .

e) Find the gradient of the curve at $P$ .

The line, $L$ , is the tangent to the graph of $f (x)$ at $P$ .

f) Find the equation of $L$ in the form $y = m x + c$ .

There is a second point, $Q$ , on the curve at which the tangent to $f (x)$ is parallel to $L$ .

g) Write down the gradient of the tangent at $Q$ .

h) Calculate the $x$ -coordinate of $Q$ .

Show answer

Answer: a) $x \approx - 1.10$ ; $x \approx 0.218$ ; $x \approx 3.13$ ; b) $f^{'} (x) = 12 x^{2} - 18 x - 12$ ; c) local maximum $6.25$ ; d) $P (0, 3)$ ; e) $- 12$ ; f) $y = - 12 x + 3$ ; g) $- 12$ ; h) $x = 1.5$

Show working and answer

We solve $4 x^{3} - 9 x^{2} - 12 x + 3 = 0$ using the GDC (or read the $x$ -intercepts from the graph): $x \approx - 1.10, x \approx 0.218, x \approx 3.13.$

b) $f^{'} (x) = 12 x^{2} - 18 x - 12.$

Extrema satisfy $f^{'} (x) = 0$ : $12 x^{2} - 18 x - 12 = 0 \Rightarrow 2 x^{2} - 3 x - 2 = 0 \Rightarrow (2 x + 1) (x - 2) = 0$ , so $x = - \frac{1}{2}$ or $x = 2$ .
On the graph, the local maximum is the one on the left, at $x = - \frac{1}{2}$ . We check this with $f^{''} (x) = 24 x - 18$ : $f^{''} (- \frac{1}{2}) = - 30 < 0$ , so it is a maximum.
We calculate its value: $f (- \frac{1}{2}) = 4 (- \frac{1}{8}) - 9 \cdot \frac{1}{4} - 12 (- \frac{1}{2}) + 3 = - 0.5 - 2.25 + 6 + 3 = 6.25.$

On the $y$ -axis $x = 0$ and $f (0) = 3$ , so $P (0, 3)$ .

The gradient of the curve at $P$ is $f^{'} (0) = - 12$ .

$L$ passes through $(0, 3)$ with gradient $- 12$ : $y = - 12 x + 3.$

Parallel tangents have the same gradient: $- 12$ .

We look for $f^{'} (x) = - 12$ : $12 x^{2} - 18 x - 12 = - 12 \Rightarrow 12 x^{2} - 18 x = 0 \Rightarrow 6 x (2 x - 3) = 0.$
One solution is $x = 0$ (the point $P$ ); the other, $x = \frac{3}{2}$ , is that of $Q$ .

Exercise 51 HardIB · P2 TZ0 November 2010

Consider the function $f (x) = x^{3} - 3 x^{2} - 24 x + 30$ .

a) Write down $f (0)$ .

b) Find $f^{'} (x)$ .

c) Find the gradient of the graph of $f (x)$ at the point where $x = 1$ .

The graph of $f (x)$ has a local maximum, $M$ , and a local minimum, $N$ .

d) (i) Use $f^{'} (x)$ to find the $x$ -coordinate of $M$ and of $N$ .

(ii) Hence, or otherwise, write down the coordinates of $M$ and of $N$ .

e) Sketch the graph of $f (x)$ for $- 5 \leq x \leq 7$ and $- 60 \leq y \leq 60$ . Clearly label $M$ and $N$ on the graph.

The lines $L_{1}$ and $L_{2}$ are parallel, and are also tangents to the graph of $f (x)$ at the points $A$ and $B$ respectively. $L_{1}$ has equation $y = 21 x + 111$ .

f) (i) Find the $x$ -coordinate of $A$ and of $B$ .

(ii) Find the $y$ -coordinate of $B$ .

Show answer

Answer: a) $30$ ; b) $f^{'} (x) = 3 x^{2} - 6 x - 24$ ; c) $- 27$ ; d) (i) $x_{M} = - 2$ , $x_{N} = 4$ ; (ii) $M (- 2, 58)$ and $N (4, - 50)$ ; e) graph (figure); f) (i) $x_{A} = - 3$ , $x_{B} = 5$ ; (ii) $y_{B} = - 40$

Show working and answer

$f (0) = 30$ .

b) $f^{'} (x) = 3 x^{2} - 6 x - 24.$

$f^{'} (1) = 3 - 6 - 24 = - 27$ .

(i) At the extrema $f^{'} (x) = 0$ : $3 x^{2} - 6 x - 24 = 0 \Rightarrow x^{2} - 2 x - 8 = 0 \Rightarrow (x - 4) (x + 2) = 0.$
Using $f^{''} (x) = 6 x - 6$ : at $x = - 2$ , $f^{''} = - 18 < 0$ (maximum, $M$ ); at $x = 4$ , $f^{''} = 18 > 0$ (minimum, $N$ ). So $x_{M} = - 2$ and $x_{N} = 4$ .
(ii) We substitute into $f$ : $f (- 2) = - 8 - 12 + 48 + 30 = 58, f (4) = 64 - 48 - 96 + 30 = - 50.$ So $M (- 2, 58)$ and $N (4, - 50)$ .

The curve rises to the maximum $M (- 2, 58)$ , falls to the minimum $N (4, - 50)$ and rises again. It crosses the $y$ -axis at $(0, 30)$ and the $x$ -axis at $x \approx - 4.27$ , $x \approx 1.15$ and $x \approx 6.12$ .

Graph of f(x) = x³ − 3x² − 24x + 30 for −5 ≤ x ≤ 7 and −60 ≤ y ≤ 60, with grid: crosses the y-axis at (0; 30); local maximum M(−2; 58) and local minimum N(4; −50) marked

(i) Parallel tangents have the same gradient. Since $L_{1}$ is $y = 21 x + 111$ , its gradient is $21$ , so $f^{'} (x) = 21$ : $3 x^{2} - 6 x - 24 = 21 \Rightarrow x^{2} - 2 x - 15 = 0 \Rightarrow (x - 5) (x + 3) = 0.$
We need to check which of the two points corresponds to $L_{1}$ . At $x = - 3$ : $f (- 3) = - 27 - 27 + 72 + 30 = 48$ and the line gives $21 (- 3) + 111 = 48$ . They coincide, so $A$ has $x$ -coordinate $- 3$ and $B$ has $x$ -coordinate $5$ .
(ii) $y$ -coordinate of $B$ : $f (5) = 125 - 75 - 120 + 30 = - 40.$

Exercise 52 HardIB · P2 TZ1 May 2009

Consider the function $f (x) = 3 x + \frac{12}{x ^{2}}$ , $x \neq = 0$ .

a) Differentiate $f (x)$ with respect to $x$ .

b) Calculate $f^{'} (x)$ when $x = 1$ .

c) Use your answer to part (b) to decide whether the function $f$ is increasing or decreasing at $x = 1$ . Justify your answer.

d) Solve the equation $f^{'} (x) = 0$ .

e) The graph of $f$ has a local minimum at the point $P$ . Let $T$ be the tangent to the graph of $f$ at $P$ .

(i) Write down the coordinates of $P$ .

(ii) Write down the gradient of $T$ .

(iii) Write down the equation of $T$ .

f) Sketch the graph of the function $f$ , for $- 3 \leq x \leq 6$ and $- 7 \leq y \leq 15$ . Clearly indicate the point $P$ and any intersections of the curve with the axes.

g) (i) On your graph, draw and label the tangent $T$ .

(ii) $T$ meets the graph of $f$ at a second point. Write down the $x$ -coordinate of this point of intersection.

Show answer

Answer: a) $f^{'} (x) = 3 - \frac{24}{x ^{3}}$ ; b) $- 21$ ; c) decreasing, since $f^{'} (1) < 0$ ; d) $x = 2$ ; e) (i) $P (2, 9)$ ; (ii) $0$ ; (iii) $y = 9$ ; f) graph (figure); g) (i) $T$ : $y = 9$ drawn; (ii) $x = - 1$

Show working and answer

We write $f (x) = 3 x + 12 x^{- 2}$ : $f^{'} (x) = 3 - 24 x^{- 3} = 3 - \frac{24}{x ^{3}} .$

$f^{'} (1) = 3 - 24 = - 21$ .

Since $f^{'} (1) = - 21 < 0$ , the function is decreasing at $x = 1$ (the gradient of the tangent is negative).

We set it equal to zero: $3 - \frac{24}{x ^{3}} = 0 \Rightarrow x^{3} = 8 \Rightarrow x = 2.$

(i) $f (2) = 6 + \frac{12}{4} = 9$ , so $P (2, 9)$ .
(ii) At a minimum the tangent is horizontal: gradient $f^{'} (2) = 0$ .
(iii) The tangent is the horizontal line through $P$ : $y = 9.$

The graph has a vertical asymptote at $x = 0$ . For $x < 0$ the curve rises to $+ \infty$ as it approaches $0$ , and crosses the $x$ -axis at $x = - 34 \approx - 1.59$ . For $x > 0$ it falls from $+ \infty$ to the minimum $P (2, 9)$ and then rises (passing through $(4; 12.75)$ ). It does not cross the $y$ -axis.

Graph of f(x)=3x+12/x² for −3 ≤ x ≤ 6 and −7 ≤ y ≤ 15, with vertical asymptote x=0, minimum P(2; 9) and horizontal tangent T

(i) $T$ is the horizontal line $y = 9$ , drawn and labelled $T$ on the graph, touching the curve at $P$ .
(ii) We look for other points with $f (x) = 9$ : $3 x + \frac{12}{x ^{2}} = 9 \Rightarrow x^{3} - 3 x^{2} + 4 = 0 \Rightarrow (x - 2)^{2} (x + 1) = 0.$ The double root $x = 2$ is $P$ ; the second point of intersection has $x = - 1$ .

Optimisation

Exercise 53 MediumIB · P2 TZ2 May 2008

The cost per person, in euros, when $x$ people are invited to a party can be determined by the function $C (x) = x + \frac{100}{x}$ .

a) Find $C^{'} (x)$ .

b) Show that the cost per person is a minimum when 10 people are invited to the party.

c) Calculate the minimum cost per person.

Show answer

Answer: a) $C^{'} (x) = 1 - \frac{100}{x ^{2}}$ ; b) minimum at $x = 10$ (shown); c) $20$ euros per person

Show working and answer

We write $C (x) = x + 100 x^{- 1}$ .
Differentiate: $C^{'} (x) = 1 - \frac{100}{x ^{2}}$ .

A minimum requires $C^{'} (x) = 0$ : $1 - \frac{100}{x ^{2}} = 0 \Rightarrow x^{2} = 100 \Rightarrow x = 10$ (only the positive root, since $x > 0$ ).
We check that it is a minimum: $C^{''} (x) = \frac{200}{x ^{3}}$ and $C^{''} (10) = 0.2 > 0$ .
Therefore, the cost per person is a minimum when $10$ people are invited.

We substitute $x = 10$ : $C (10) = 10 + \frac{100}{10} = 20.$
The minimum cost per person is $20$ euros.

Answer: a) $C^{'} (x) = 1 - \frac{100}{x ^{2}}$ ; b) minimum at $x = 10$ (shown); c) $20$ euros per person

Exercise 54 MediumIB · P2 TZ1 May 2018

A manufacturer makes rubbish bins in the shape of a cylinder with a hemispherical lid.

The rubbish bin has a height of 70 cm. The base radius of both the cylinder and the hemispherical top is 20 cm.

Rubbish bin seen from the front: vertical cylinder with a hemispherical lid, with the word Basura on the lid, a vertical dimension of 70 cm from the base to the top of the lid, and a base radius of 20 cm

a) Write down the height of the cylinder.

b) Find the total volume of the rubbish bin.

A designer is asked to produce a new rubbish bin.

The new rubbish bin will also be in the shape of a cylinder with a hemispherical lid.

This rubbish bin will have a height of $H$ cm and a base radius of $r$ cm.

Rubbish bin seen from the front, with vertical dimension H from the base to the top of the lid and base radius r

There is a design constraint such that $H + 2 r = 110$ cm.

The designer has to maximise the volume of the rubbish bin.

c) Find the height of the cylinder, $h$ , of the new rubbish bin, in terms of $r$ .

d) Show that the volume, $V$ cm $^{3}$ , of the new rubbish bin is given by $V = 110 π r^{2} - \frac{7}{3} π r^{3}$ .

e) Using your graphic display calculator, find the value of $r$ that maximises the value of $V$ .

A designer claims that the new rubbish bin has a capacity that is at least 40 % greater than the capacity of the original rubbish bin.

f) State whether the designer's claim is correct. Justify your answer.

Show answer

Answer: a) $50$ cm; b) $\frac{76 000}{3} π \approx 79587 cm^{3}$ ; c) $h = 110 - 3 r$ ; d) shown; e) $r \approx 31.4$ cm; f) correct: the maximum volume ( $\approx 113800 cm^{3}$ ) is $43 %$ greater

Show working and answer

The hemispherical lid takes up $20$ cm of the total height (its radius), so the height of the cylinder is $70 - 20 = 50$ cm.

Cylinder: $π \cdot 2 0^{2} \cdot 50 = 20000 π$ cm $^{3}$ .
Hemisphere: $\frac{1}{2} \cdot \frac{4}{3} π \cdot 2 0^{3} = \frac{16 000}{3} π$ cm $^{3}$ .
Total volume: $V = 20000 π + \frac{16 000}{3} π = \frac{76 000}{3} π \approx 79600 cm^{3} (79587 cm^{3}) .$

The total height is $H = h + r$ , and $H = 110 - 2 r$ . Then $h = H - r = (110 - 2 r) - r = 110 - 3 r .$

$V = π r^{2} h + \frac{2}{3} π r^{3} = π r^{2} (110 - 3 r) + \frac{2}{3} π r^{3}$ .
Expanding: $V = 110 π r^{2} - 3 π r^{3} + \frac{2}{3} π r^{3} = 110 π r^{2} - \frac{7}{3} π r^{3} .$

Using the GDC, the maximum of the graph of $V (r)$ occurs at $r \approx 31.4$ cm.
Check: $V^{'} (r) = 220 π r - 7 π r^{2} = π r (220 - 7 r) = 0$ gives $r = \frac{220}{7} \approx 31.4$ cm.

With $r = \frac{220}{7}$ cm: $V = 110 π r^{2} - \frac{7}{3} π r^{3} \approx 113800 cm^{3}$ .
We compare with the original bin: $\frac{113 781}{79 587} \approx 1.43.$
The new volume is $43 %$ greater, which exceeds $40 %$ : the designer's claim is correct.

Exercise 55 MediumIB · P2 TZ2 May 2012

A container for transporting goods, with six rectangular faces, is to be constructed, as shown in the diagram.

Rectangular box in perspective with grey left side face: the bottom edge of that face is labelled x (width), the bottom edge of the front face 2x (length) and the right vertical edge of the front face y (height); no numerical scale

The dimensions of the container are as follows: length $2 x$ , width $x$ , height $y$ .

All measurements are in metres. The total length of the twelve edges of the container is 48 metres.

a) Show that $y = 12 - 3 x$ .

b) Show that the volume of the container $V$ m $^{3}$ is given by $V = 24 x^{2} - 6 x^{3}$ .

c) Find $\frac{d V}{d x}$ .

d) Find the value of $x$ for which $V$ is a maximum.

e) Find the maximum volume of the container.

f) Find the length and the height of the container for which the volume is a maximum.

The container is to be painted. One litre of paint covers an area of 15 m $^{2}$ . The paint is sold in tins of four litres.

g) Calculate the number of tins needed to paint the entire container.

Show answer

Answer: a) shown; b) shown; c) $\frac{d V}{d x} = 48 x - 18 x^{2}$ ; d) $x = \frac{8}{3} \approx 2.67$ m; e) $V_{m a x} = \frac{512}{9} \approx 56.9$ m $^{3}$ ; f) length $\approx 5.33$ m and height $4$ m; g) $2$ tins

Show working and answer

The container has $4$ edges of length $2 x$ , $4$ of length $x$ and $4$ of length $y$ : $4 (2 x) + 4 x + 4 y = 48 \Rightarrow 12 x + 4 y = 48.$
Dividing by $4$ : $3 x + y = 12$ , i.e. $y = 12 - 3 x$ .

$V = 2 x \cdot x \cdot y = 2 x^{2} (12 - 3 x) = 24 x^{2} - 6 x^{3}$ .

c) $\frac{d V}{d x} = 48 x - 18 x^{2} .$

$48 x - 18 x^{2} = 0 \Rightarrow 6 x (8 - 3 x) = 0$ .
Since $x > 0$ : $x = \frac{8}{3} \approx 2.67$ m.
It is a maximum: $\frac{d ^{2} V}{d x ^{2}} = 48 - 36 x = - 48 < 0$ at $x = \frac{8}{3}$ .

e) $V = 24 (\frac{8}{3})^{2} - 6 (\frac{8}{3})^{3} = \frac{512}{3} - \frac{1024}{9} = \frac{512}{9} \approx 56.9 m^{3} .$

Length: $2 x = \frac{16}{3} \approx 5.33$ m.
Height: $y = 12 - 3 \cdot \frac{8}{3} = 4$ m.

Total surface area: two faces of $2 x \cdot x$ , two of $2 x \cdot y$ and two of $x \cdot y$ : $S = 2 (2 x^{2} + 2 x y + x y) = 4 x^{2} + 6 x y = 4 \cdot \frac{64}{9} + 6 \cdot \frac{8}{3} \cdot 4 \approx 92.4 m^{2} .$
Paint required: $\frac{92.4}{15} \approx 6.16$ litres.
Each tin contains $4$ litres: one tin is not enough (4 L), but two are (8 L).

Exercise 56 HardIB · P2 TZ0 November 2019

The Maxwell Ohm Company is designing a portable Bluetooth speaker. The speaker is in the shape of a cylinder with a hemisphere at each end of the cylinder.

Elongated portable speaker seen from the front: body shaped like a capsule, narrow grille on the top and a circular speaker at each end

The following figure shows the dimensions of the speaker in centimetres; $r$ is the radius of the hemisphere and $l$ is the length of the cylinder, where $r > 0$ and $l \geq 0$ .

Side view of the speaker: horizontal cylinder with a semicircle at each end, two dashed lines at the joins, dimension l of the cylindrical section and radius r of the left semicircle

a) Write down an expression for $V$ , the volume of the speaker in $cm^{3}$ , in terms of $r$ , $l$ and $π$ .

The Maxwell Ohm Company has decided that the speaker will have a surface area of $300 cm^{2}$ .

b) Write down an equation that allows the surface area of the speaker to be calculated in terms of $r$ , $l$ and $π$ .

c) Given that the design must satisfy $l = \frac{150 - 2 π r ^{2}}{π r}$ , show that $V = 150 r - \frac{2 π r ^{3}}{3} .$

d) Find $\frac{d V}{d r}$ .

The sound quality of the speaker improves as $V$ increases.

e) Using your answer to part (d), show that $V$ is a maximum when $r$ is equal to $\frac{75}{π}$ cm.

f) Find the length of the cylinder for which $V$ is a maximum.

g) Calculate the maximum value of $V$ .

h) Use your answer to part (f) to identify the shape of the speaker that gives the best sound quality.

Show answer

Answer: a) $V = π r^{2} l + \frac{4}{3} π r^{3}$ ; b) $2 π r l + 4 π r^{2} = 300$ ; c) shown; d) $\frac{d V}{d r} = 150 - 2 π r^{2}$ ; e) shown; f) $l = 0$ cm; g) $V \approx 489 cm^{3}$ ; h) a sphere of radius $75/ π \approx 4.89$ cm

Show working and answer

The two hemispheres together form a sphere of radius $r$ . The volume is that of the cylinder plus that of the sphere: $V = π r^{2} l + \frac{4}{3} π r^{3} .$

The surface area is the lateral area of the cylinder ( $2 π r l$ ) plus that of the sphere ( $4 π r^{2}$ ). Setting this equal to $300$ : $2 π r l + 4 π r^{2} = 300.$

We substitute $l = \frac{150 - 2 π r ^{2}}{π r}$ into the volume: $V = π r^{2} \cdot \frac{150 - 2 π r ^{2}}{π r} + \frac{4}{3} π r^{3} = r (150 - 2 π r^{2}) + \frac{4}{3} π r^{3} .$
We expand and collect terms: $V = 150 r - 2 π r^{3} + \frac{4}{3} π r^{3} = 150 r - \frac{2 π r ^{3}}{3} .$

$\frac{d V}{d r} = 150 - 2 π r^{2}$ .

The maximum occurs where $\frac{d V}{d r} = 0$ : $150 - 2 π r^{2} = 0 \Rightarrow r^{2} = \frac{75}{π} \Rightarrow r = \frac{75}{π} cm$ (only the positive root, since $r > 0$ ).
The second derivative is $\frac{d ^{2} V}{d r ^{2}} = - 4 π r < 0$ , so it is a maximum.

With $r^{2} = \frac{75}{π}$ we get $2 π r^{2} = 150$ , so $l = \frac{150 - 150}{π r} = 0 cm .$

With $l = 0$ the volume is $V = 150 r - \frac{2 π r ^{3}}{3}$ . Since $π r^{2} = 75$ , this becomes $V = 150 r - 50 r = 100 r$ .
Substituting: $V = 100 \frac{75}{π} \approx 489 cm^{3} .$

With $l = 0$ there is no cylindrical part: the speaker with the best sound quality is a sphere of radius $\frac{75}{π} \approx 4.89$ cm.

Exercise 57 HardIB · P2 TZ2 May 2016

Hugo is given a rectangular piece of thin card measuring 16 cm by 10 cm. He decides to use it to design a tray.

From the corners he cuts out the shaded squares, of side $x$ cm, as shown in the following diagram.

Rectangular card 16 cm by 10 cm with a shaded square of side x at each corner, marking the cuts

He folds what remains of the card to make a tray, as shown in the following diagram.

Open tray in perspective: a horizontal rectangular base and four low walls, with no lid

a) Write down, in terms of $x$ , the length and the width of the tray.

b) (i) State whether the value of $x$ can be equal to 5. Give a reason to justify your answer.

(ii) Write down the interval of possible values of $x$ .

c) Show that the volume, $V$ cm $^{3}$ , of this tray is given by $V = 4 x^{3} - 52 x^{2} + 160 x$ .

d) Find $\frac{d V}{d x}$ .

e) Using your answer to part (d), find the value of $x$ that maximises the volume of the tray.

f) Calculate the maximum volume of the tray.

g) Sketch the graph of $V = 4 x^{3} - 52 x^{2} + 160 x$ for the possible values of $x$ found in part (b)(ii), and $0 \leq V \leq 200$ . Clearly label the maximum point.

Show answer

Answer: a) length $16 - 2 x$ cm and width $10 - 2 x$ cm; b) (i) no, the width would be $0$ ; (ii) $0 < x < 5$ ; c) shown; d) $\frac{d V}{d x} = 12 x^{2} - 104 x + 160$ ; e) $x = 2$ cm; f) $V_{m a x} = 144 cm^{3}$ ; g) curve from $(0, 0)$ to $(5, 0)$ with the maximum labelled at $(2, 144)$

Show working and answer

When a square of side $x$ is cut from each corner, each side of the card loses $x$ at each end, that is $2 x$ in total.
Length of the tray: $16 - 2 x$ cm. Width: $10 - 2 x$ cm.

(i) No. With $x = 5$ the width would be $10 - 2 \cdot 5 = 0$ , and there would be no tray.
(ii) We need $x > 0$ , $10 - 2 x > 0$ and $16 - 2 x > 0$ . The most restrictive condition is $x < 5$ , so $0 < x < 5.$

The volume is the area of the base times the height (which is $x$ ): $V = (16 - 2 x) (10 - 2 x) x .$
Expand the product: $(16 - 2 x) (10 - 2 x) = 160 - 32 x - 20 x + 4 x^{2} = 4 x^{2} - 52 x + 160.$
Multiplying by $x$ : $V = 4 x^{3} - 52 x^{2} + 160 x .$

Differentiate: $\frac{d V}{d x} = 12 x^{2} - 104 x + 160.$

Set equal to zero and simplify: $12 x^{2} - 104 x + 160 = 0 \Rightarrow 3 x^{2} - 26 x + 40 = 0.$
Apply the quadratic formula: $x = \frac{26 \pm 676 - 480}{6} = \frac{26 \pm 14}{6} \Rightarrow x = 2 or x = \frac{20}{3} .$
Only $x = 2$ lies in the interval $0 < x < 5$ .
It is a maximum because $\frac{d ^{2} V}{d x ^{2}} = 24 x - 104$ equals $- 56 < 0$ at $x = 2$ .

Substitute $x = 2$ : $V (2) = 4 \cdot 8 - 52 \cdot 4 + 160 \cdot 2 = 32 - 208 + 320 = 144.$
The maximum volume is $144 cm^{3}$ .

Sketch $V (x)$ for $0 < x < 5$ with $0 \leq V \leq 200$ : the curve starts at $(0, 0)$ , rises to the maximum $(2, 144)$ and falls to cut the horizontal axis at $(5, 0)$ .
The maximum point is labelled with its coordinates $(2, 144)$ .

Graph of V against x for 0 ≤ x ≤ 5 and 0 ≤ V ≤ 200: rises from the origin to the maximum (2; 144) and falls to (5; 0)

Exercise 58 HardIB · P2 TZ0 November 2016

A water container is in the shape of a cylinder. The internal height is $h$ cm and the internal radius of the base is $r$ cm.

Vertical cylinder open at the top, with internal height h and internal radius r at the base

The water container has no lid. The internal surfaces of the container are going to be coated with a waterproofing material.

a) Write down a formula that gives the area, $A$ , of the surface to be coated.

The volume of the water container is equal to 0.5 m $^{3}$ .

b) Express this volume in cm $^{3}$ .

c) Write down, in terms of $r$ and $h$ , an equation that gives the volume of this water container.

d) Show that $A = π r^{2} + \frac{1000000}{r}$ .

The water container is designed so that the area to be coated is minimised.

e) Find $\frac{d A}{d r}$ .

f) Using your answer to part (e), find the value of $r$ that minimises $A$ .

g) Find the value of this minimum area.

A tin of waterproofing material can coat an area of 2000 cm $^{2}$ .

h) Find the minimum number of tins of waterproofing material needed to coat the whole area calculated in part (g).

Show answer

Answer: a) $A = 2 π r h + π r^{2}$ ; b) $500000 cm^{3}$ ; c) $π r^{2} h = 500000$ ; d) shown; e) $\frac{d A}{d r} = 2 π r - \frac{1 000 000}{r ^{2}}$ ; f) $r \approx 54.2$ cm; g) $A \approx 27700 cm^{2}$ ; h) $14$ tins

Show working and answer

The lateral surface (a rectangle of $2 π r$ by $h$ ) and the base (a circle) are coated, with no lid: $A = 2 π r h + π r^{2} .$

Since $1 m^{3} = 1 0^{6} cm^{3}$ : $0.5 m^{3} = 0.5 \cdot 1 0^{6} = 500000 cm^{3} .$

The volume of a cylinder is the area of the base times the height: $π r^{2} h = 500 000.$

Solve the previous equation for the height: $h = \frac{500 000}{π r ^{2}} .$
Substitute into $A$ : $A = 2 π r \cdot \frac{500 000}{π r ^{2}} + π r^{2} = \frac{1 000 000}{r} + π r^{2},$ which is what was asked: $A = π r^{2} + \frac{1 000 000}{r}$ .

Write $A = π r^{2} + 1 0^{6} r^{- 1}$ and differentiate: $\frac{d A}{d r} = 2 π r - \frac{1 000 000}{r ^{2}} .$

Set equal to zero: $2 π r - \frac{1 000 000}{r ^{2}} = 0 \Rightarrow 2 π r^{3} = 1000000 \Rightarrow r^{3} = \frac{500 000}{π} .$
Then $r = 3 \frac{500 000}{π} \approx 54.2 cm .$
It is a minimum because $\frac{d ^{2} A}{d r ^{2}} = 2 π + \frac{2 000 000}{r ^{3}} > 0$ .

Substitute $r = 54.19 \dots$ into $A$ : $A = π (54.19)^{2} + \frac{1 000 000}{54.19} \approx 27679 cm^{2} \approx 27700 cm^{2} .$

Each tin covers $2000 cm^{2}$ : $\frac{27 679}{2000} \approx 13.8.$
As fractions of a tin cannot be used, $14$ tins are needed.

Exercise 59 HardIB · P2 TZ1 May 2014

A lobster trap is made in the shape of a half-cylinder. It is built from a steel frame with nets pulled tightly around it. The steel frame consists of a rectangular base, two semicircular ends and two additional support bars, as shown in the following diagram.

Lobster trap in the shape of a horizontal half-cylinder: rectangular base, two semicircular ends and two top bars; radius r marked from the centre of the end and length l of the long side; a lobster is drawn on top

The semicircular ends have radius $r$ and each support bar has length $l$ .

Let $T$ be the total length of steel used in the frame of the lobster trap.

a) Write an expression in terms of $r$ , $l$ and $π$ .

The volume of the lobster trap is 0.75 m $^{3}$ .

b) Write an equation for the volume of the lobster trap in terms of $r$ , $l$ and $π$ .

c) Show that $T = (2 π + 4) r + \frac{6}{π r ^{2}}$ .

d) Calculate $\frac{d T}{d r}$ .

The lobster trap is designed so that the length of steel used in its frame is a minimum.

e) Show that the value of $r$ for which $T$ is a minimum is 0.719 m, correct to three significant figures.

f) Calculate the value of $l$ for which $T$ is a minimum.

g) Calculate the minimum value of $T$ .

Show answer

Answer: a) $T = 4 l + 4 r + 2 π r$ ; b) $\frac{1}{2} π r^{2} l = 0.75$ ; c) shown; d) $\frac{d T}{d r} = 2 π + 4 - \frac{12}{π r ^{3}}$ ; e) $r \approx 0.719$ m; f) $l \approx 0.924$ m; g) $T_{m i n} \approx 11.1$ m

Show working and answer

Rectangular base: two sides of length $l$ and two sides of length $2 r$ (the diameter of each semicircle).
The two support bars total $2 l$ and the two semicircular ends total $2 π r$ .
Adding everything: $T = 2 l + 2 (2 r) + 2 l + 2 π r = 4 l + 4 r + 2 π r .$

The volume is that of a half-cylinder of radius $r$ and length $l$ : $\frac{1}{2} π r^{2} l = 0.75.$

From (b): $π r^{2} l = 1.5$ , so $l = \frac{1.5}{π r ^{2}}$ .
Substitute into $T$ : $T = 4 \cdot \frac{1.5}{π r ^{2}} + 4 r + 2 π r = \frac{6}{π r ^{2}} + (2 π + 4) r .$

Write $T = (2 π + 4) r + \frac{6}{π} r^{- 2}$ and differentiate: $\frac{d T}{d r} = 2 π + 4 - \frac{12}{π r ^{3}} .$

At the minimum, $\frac{d T}{d r} = 0$ : $2 π + 4 = \frac{12}{π r ^{3}} \Rightarrow r^{3} = \frac{12}{π ( 2 π + 4 )} .$
Then $r = (\frac{12}{π ( 2 π + 4 )})^{1/3} = 0.7188 \dots \approx 0.719$ m (3 sf).
It is a minimum because $\frac{d ^{2} T}{d r ^{2}} = \frac{36}{π r ^{4}} > 0$ .

With $r = 0.7188$ : $l = \frac{1.5}{π ( 0.7188 ) ^{2}} \approx 0.924 m .$

Substitute into $T$ : $T = (2 π + 4) (0.7188) + \frac{6}{π ( 0.7188 ) ^{2}} \approx 7.39 + 3.70 \approx 11.1 m .$

Exercise 60 HardIB · P2 TZ2 May 2014

A package is in the shape of a rectangular prism, as shown in the following diagram.

It has a length of $l$ cm, a width of $w$ cm and a height of 20 cm.

The total volume of the package is equal to $3000$ cm $^{3}$ .

Package in the shape of a rectangular prism 20 cm high, with a vertical string around the section and a horizontal string at half height; hidden edges dashed

a) Express the volume of the package in terms of $l$ and $w$ .

b) Show that $l = \frac{150}{w}$ .

A piece of string that is just long enough to go once around the whole package is used to tie it, as shown in the figure.

c) Show that the length of string, $S$ cm, needed to tie the package can be written as $S = 40 + 4 w + \frac{300}{w}$ , $0 < w \leq 20$ .

d) Draw accurately the graph of $S$ for $0 < w \leq 20$ and $0 < S \leq 500$ , clearly showing the local minimum. Use a scale of 2 cm to represent 5 units on the horizontal $w$ (cm) axis and 2 cm to represent 100 units on the vertical $S$ (cm) axis.

e) Find $\frac{d S}{d w}$ .

f) Find the value of $w$ for which $S$ is a minimum.

g) Write down the value of $l$ , the length of the package, for which the length of string is a minimum.

h) Find the minimum length of string needed to tie the package.

Show answer

Answer: a) $V = 20 l w$ ; b) shown; c) shown; d) graph with local minimum at $(8.66; 109)$ ; e) $\frac{d S}{d w} = 4 - \frac{300}{w ^{2}}$ ; f) $w = 53 \approx 8.66$ cm; g) $l \approx 17.3$ cm; h) $S_{m i n} = 40 + 403 \approx 109$ cm

Show working and answer

Volume of a prism: length × width × height. $V = l \cdot w \cdot 20 = 20 l w .$

Equate to the given volume: $20 l w = 3000$ .
Divide by $20$ : $l w = 150$ , so $l = \frac{150}{w} .$

The vertical string goes around the section of width $w$ and height $20$ : length $2 w + 40$ .
The horizontal string goes around the package along $l$ and $w$ : length $2 l + 2 w$ .
Adding and substituting $l = \frac{150}{w}$ : $S = (2 w + 40) + (2 l + 2 w) = 40 + 4 w + 2 \cdot \frac{150}{w} = 40 + 4 w + \frac{300}{w} .$

$S = 40 + 4 w + \frac{300}{w}$ is plotted on the GDC, with $0 < w \leq 20$ and $0 < S \leq 500$ .
The curve has the vertical asymptote $w = 0$ (upwards, $S$ grows without bound as $w \to 0$ ), decreases to a local minimum at $(8.66; 109)$ and rises again to $S (20) = 40 + 80 + 15 = 135$ .
On paper: $2$ cm represent $5$ cm in $w$ and $2$ cm represent $100$ cm in $S$ .

Graph of S = 4w + 40 + 300/w for 0 < w ≤ 20: vertical asymptote at w = 0, local minimum (8.66; 109) and endpoint (20; 135)

Write $S = 40 + 4 w + 300 w^{- 1}$ and differentiate: $\frac{d S}{d w} = 4 - \frac{300}{w ^{2}} .$

At the minimum, $\frac{d S}{d w} = 0$ : $4 = \frac{300}{w ^{2}} \Rightarrow w^{2} = 75 \Rightarrow w = 53 \approx 8.66 cm .$
Only the positive value is valid. It is a minimum because $\frac{d ^{2} S}{d w ^{2}} = \frac{600}{w ^{3}} > 0$ .

$l = \frac{150}{w} = \frac{150}{5 3} = 103 \approx 17.3$ cm.

Substitute $w = 53$ into $S$ : $S = 40 + 4 \cdot 53 + \frac{300}{5 3} = 40 + 203 + 203 = 40 + 403 \approx 109 cm .$

Exercise 61 HardIB · P2 TZ0 November 2014

Tipis were traditionally used by nomadic tribes who lived on the Great Plains of North America. They are conical dwellings and can be modelled as a cone, with vertex O, shown below. The cone has radius $r$ , height $h$ and slant height $l$ .

Two cones in perspective: on the left, a traditional tipi with a dark zigzag band; on the right, a right cone with vertex O, the height h dashed, the slant edge l oblique and the radius r of the base marked with a right angle

A model tipi is on display at a Great Plains exhibition. The curved surface of this tipi is covered by canvas of area 39.27 m $^{2}$ , which has the shape of a semicircle, as shown in the following diagram.

Right cone with vertex O, height h dashed, slant height l and base radius r marked with a right angle; on the right, a shaded semicircular surface of radius l and area 39.27 m²

a) Show that the slant height, $l$ , is 5 m, correct to the nearest metre.

b) (i) Find the circumference of the base of the cone.

(ii) Find the radius, $r$ , of the base.

(iii) Find the height, $h$ .

A company designs cone-shaped tents to resemble traditional tipis.

These cone-shaped tents come in a variety of sizes such that the sum of the diameter and the height is equal to 9.33 m.

c) Write an expression for the height, $h$ , in terms of the radius, $r$ , of these cone-shaped tents.

d) Show that the volume of the tent, $V$ , can be written as $V = 3.11 π r^{2} - \frac{2}{3} π r^{3}$ .

e) Calculate $\frac{d V}{d r}$ .

f) (i) Determine the exact value of $r$ for which the volume is a maximum.

(ii) Find the maximum volume.

Show answer

Answer: a) $l = 5$ m; b) (i) $5 π \approx 15.7$ m; (ii) $r = 2.5$ m; (iii) $h \approx 4.33$ m; c) $h = 9.33 - 2 r$ ; d) shown; e) $\frac{d V}{d r} = 6.22 π r - 2 π r^{2}$ ; f) (i) $r = 3.11$ m; (ii) $V_{m a x} \approx 31.5$ m $^{3}$

Show working and answer

The canvas is a semicircle of radius $l$ , so its area is $\frac{1}{2} π l^{2}$ : $\frac{1}{2} π l^{2} = 39.27 \Rightarrow l^{2} = \frac{2 \cdot 39.27}{π} = 25.0 \dots$
Therefore $l = 25 = 5$ m (to the nearest metre).

(i) The circumference of the base coincides with the arc length of the semicircle, which is half the circumference of radius $5$ : $π \cdot 5 = 5 π \approx 15.7 m .$
(ii) $2 π r = 5 π \Rightarrow r = 2.5$ m.
(iii) By Pythagoras, $h^{2} + r^{2} = l^{2}$ : $h = 25 - 6.25 = 18.75 \approx 4.33 m .$

The diameter plus the height equals $9.33$ : $2 r + h = 9.33$ .
Solving for $h$ : $h = 9.33 - 2 r .$

Volume of a cone: $V = \frac{1}{3} π r^{2} h$ .
Substitute $h$ : $V = \frac{1}{3} π r^{2} (9.33 - 2 r) = 3.11 π r^{2} - \frac{2}{3} π r^{3} .$

e) $\frac{d V}{d r} = 6.22 π r - 2 π r^{2} .$

(i) Set equal to zero: $π r (6.22 - 2 r) = 0 \Rightarrow r = 3.11 m (r = 0 is not meaningful) .$ It is a maximum because $\frac{d ^{2} V}{d r ^{2}} = 6.22 π - 4 π r = - 6.22 π < 0$ at $r = 3.11$ .
(ii) With $r = 3.11$ , $h = 9.33 - 6.22 = 3.11$ : $V = \frac{1}{3} π (3.11)^{2} (3.11) = \frac{π ( 3.11 ) ^{3}}{3} \approx 31.5 m^{3} .$

Exercise 62 HardIB · P2 TZ2 May 2010

A dog food manufacturer wants to reduce its production costs. When manufacturing the cylindrical cans, it wants to use as little aluminium as possible. In the figure below, $h$ represents the height of the can in cm and $x$ the radius of the base of the can in cm.

Vertical cylindrical can in perspective, with the height h and the radius x marked, no numerical scale

The volume of the dog food cans is equal to $600 cm^{3}$ .

a) Verify that $h = \frac{600}{π x ^{2}}$ .

b) (i) Find an expression for the curved surface area of the can, in terms of $x$ . Simplify your answer.

(ii) Hence, write down an expression for the total surface area of the can, $A$ , in terms of $x$ .

c) Differentiate $A$ with respect to $x$ .

d) Find the value of $x$ for which $A$ is a minimum.

e) Calculate the minimum total surface area of the dog food cans.

Show answer

Answer: a) $h = \frac{600}{π x ^{2}}$ ; b) (i) $\frac{1200}{x}$ ; (ii) $A = \frac{1200}{x} + 2 π x^{2}$ ; c) $\frac{d A}{d x} = - \frac{1200}{x ^{2}} + 4 π x$ ; d) $x \approx 4.57$ cm; e) $A_{m i n} \approx 394 cm^{2}$

Show working and answer

The volume of the cylinder is $V = π x^{2} h = 600$ .
We solve for $h$ : $h = \frac{600}{π x ^{2}} .$

(i) The curved surface is a rectangle with base $2 π x$ (the circumference) and height $h$ : $2 π x \cdot \frac{600}{π x ^{2}} = \frac{1200}{x} .$
(ii) We add the two circular bases, each of area $π x^{2}$ : $A = \frac{1200}{x} + 2 π x^{2} .$

We write $A = 1200 x^{- 1} + 2 π x^{2}$ and differentiate: $\frac{d A}{d x} = - \frac{1200}{x ^{2}} + 4 π x .$

At the minimum $\frac{d A}{d x} = 0$ : $4 π x = \frac{1200}{x ^{2}} \Rightarrow x^{3} = \frac{300}{π} \Rightarrow x = 3 \frac{300}{π} \approx 4.57 cm .$
It is a minimum because $\frac{d ^{2} A}{d x ^{2}} = \frac{2400}{x ^{3}} + 4 π > 0$ for $x > 0$ .

We substitute into $A$ : $A = \frac{1200}{4.5708} + 2 π (4.5708)^{2} \approx 262.5 + 131.3 \approx 394 cm^{2} .$

Exercise 63 HardIB · P2 TZ1 May 2008

A closed rectangular box has height $y$ cm and width $x$ cm. Its length is twice its width. It has a fixed external surface area of $300 cm^{2}$ .

Closed rectangular box in perspective, with the front face of width 2x, depth x and height y, and the right side face shaded

a) Show that $4 x^{2} + 6 x y = 300$ .

b) Find an expression for $y$ in terms of $x$ .

c) Using the above, show that the volume $V$ of the box is given by $V = 100 x - \frac{4}{3} x^{3} .$

d) Calculate $\frac{d V}{d x}$ .

e) (i) Using the above, find the value of $x$ and of $y$ required to maximise the volume of the box.

(ii) Calculate the maximum volume.

Show answer

Answer: a) $4 x^{2} + 6 x y = 300$ ; b) $y = \frac{150 - 2 x ^{2}}{3 x}$ ; c) shown; d) $\frac{d V}{d x} = 100 - 4 x^{2}$ ; e) (i) $x = 5$ cm, $y = \frac{20}{3} \approx 6.67$ cm; (ii) $V_{m a x} = \frac{1000}{3} \approx 333 cm^{3}$

Show working and answer

The top and bottom faces measure $2 x$ by $x$ : two faces of area $2 x^{2}$ , so $4 x^{2}$ in total.
The front and back faces measure $2 x$ by $y$ : in total $2 \cdot 2 x y = 4 x y$ .
The two side faces measure $x$ by $y$ : in total $2 x y$ .
The surface area is $4 x^{2} + 4 x y + 2 x y = 4 x^{2} + 6 x y = 300.$

We solve for $y$ : $y = \frac{300 - 4 x ^{2}}{6 x} = \frac{150 - 2 x ^{2}}{3 x} .$

The volume is length times width times height: $V = 2 x \cdot x \cdot y = 2 x^{2} y$ .
We substitute $y$ and simplify: $V = 2 x^{2} \cdot \frac{150 - 2 x ^{2}}{3 x} = \frac{2 x ( 150 - 2 x ^{2} )}{3} = 100 x - \frac{4}{3} x^{3} .$

d) $\frac{d V}{d x} = 100 - 4 x^{2} .$

(i) At the maximum $\frac{d V}{d x} = 0$ : $4 x^{2} = 100$ , so $x = 5$ cm (we reject $x = - 5$ ). It is a maximum because $\frac{d ^{2} V}{d x ^{2}} = - 8 x = - 40 < 0$ .
We calculate $y$ : $y = \frac{150 - 2 \cdot 25}{3 \cdot 5} = \frac{100}{15} = \frac{20}{3} \approx 6.67 cm .$
(ii) The maximum volume is $V = 100 \cdot 5 - \frac{4}{3} \cdot 125 = 500 - \frac{500}{3} = \frac{1000}{3} \approx 333 cm^{3} .$

Exercise 64 HardIB · P2 TZ0 November 2006

A farmer has a rectangular enclosure with a straight hedge running down one side. The area of the enclosure is $162 m^{2}$ . He encloses this area using $x$ metres of the hedge on one side, as shown in the following diagram.

Rectangular enclosure with a hedge of length x on the right side and fencing on the other three sides; diagram not to scale

a) If he uses $y$ metres of fencing to complete the enclosure, show that $y = x + \frac{324}{x}$ .

The farmer wishes to use the least amount of fencing.

b) Find $\frac{d y}{d x}$ .

c) Find the value of $x$ that makes $y$ a minimum.

d) Calculate this minimum value of $y$ .

e) Using $y = x + \frac{324}{x}$ , find the values of $a$ and $b$ in the following table.

$x$	6	9	12	18	24	27	36
$y$	60	45	39	$a$	37.5	$b$	45

f) Draw an accurate graph of this function using a horizontal scale starting at $0$ , with $2$ cm representing $10$ metres, and a vertical scale starting at $30$ , with $4$ cm representing $10$ metres.

g) Write down the values of $x$ for which $y$ is increasing.

Show answer

Answer: a) shown; b) $\frac{d y}{d x} = 1 - \frac{324}{x ^{2}}$ ; c) $x = 18$ m; d) $y_{m i n} = 36$ m; e) $a = 36$ , $b = 39$ ; f) graph (figure); g) $x > 18$

Show working and answer

Let $w$ be the other side of the rectangle. The area is $x w = 162$ , so $w = \frac{162}{x}$ .
The three sides that are not hedge are fenced: one side of length $x$ (opposite the hedge) and two of length $w$ . Then $y = x + 2 w = x + \frac{324}{x} .$

b) $\frac{d y}{d x} = 1 - \frac{324}{x ^{2}} .$

At the minimum $\frac{d y}{d x} = 0$ : $x^{2} = 324$ , so $x = 18$ m.
It is a minimum because $\frac{d ^{2} y}{d x ^{2}} = \frac{648}{x ^{3}} > 0$ .

d) $y (18) = 18 + \frac{324}{18} = 18 + 18 = 36 m .$

$a = y (18) = 36$ .
$b = y (27) = 27 + \frac{324}{27} = 27 + 12 = 39$ .

We plot the points of the table $(6, 60), (9, 45), (12, 39), (18, 36), (24, 37.5), (27, 39), (36, 45)$ , with $x$ from $0$ ( $2$ cm per $10$ m) and $y$ from $30$ ( $4$ cm per $10$ m), and join them with a smooth U-shaped curve with minimum at $(18, 36)$ .

Graph of y=x+324/x for 6 ≤ x ≤ 36 with a scale of 2 cm per 10 m horizontally and 4 cm per 10 m vertically; minimum at (18; 36)

The function increases to the right of the minimum: $y$ is increasing for $x > 18$ .

Answer: a) shown; b) $\frac{d y}{d x} = 1 - \frac{324}{x ^{2}}$ ; c) $x = 18$ m; d) $y_{m i n} = 36$ m; e) $a = 36$ , $b = 39$ ; f) graph (figure); g) $x > 18$

Antiderivatives and indefinite integrals

Exercise 65 BasicIB · P1 TZ0 November 2015

Let $f^{'} (x) = 6 x^{2} - 5$ . Given that $f (2) = - 3$ , find $f (x)$ .

Show answer

Answer: $f (x) = 2 x^{3} - 5 x - 9$

Show working and answer

We integrate $f^{'} (x)$ : $f (x) = \int (6 x^{2} - 5) d x = 2 x^{3} - 5 x + C .$
We use $f (2) = - 3$ : $2 \cdot 2^{3} - 5 \cdot 2 + C = - 3 \Rightarrow 16 - 10 + C = - 3 \Rightarrow C = - 9.$

Answer: $f (x) = 2 x^{3} - 5 x - 9$

Exercise 66 BasicIB · P1 TZ1 May 2018

A function $f (x)$ has derivative $f^{'} (x) = 3 x^{2} + 18 x$ . The graph of $f$ intercepts the $x$ -axis at $x = - 1$ .

Find $f (x)$ .

Show answer

Answer: $f (x) = x^{3} + 9 x^{2} - 8$

Show working and answer

We integrate $f^{'} (x)$ : $f (x) = \int (3 x^{2} + 18 x) d x = x^{3} + 9 x^{2} + C .$
The graph crossing the $x$ -axis at $x = - 1$ means $f (- 1) = 0$ : $(- 1)^{3} + 9 (- 1)^{2} + C = 0 \Rightarrow - 1 + 9 + C = 0 \Rightarrow C = - 8.$

Answer: $f (x) = x^{3} + 9 x^{2} - 8$

Exercise 67 BasicIB · P1 TZ2 May 2007

Let $f^{'} (x) = 12 x^{2} - 2$ . Given that $f (- 1) = 1$ , find $f (x)$ .

Show answer

Answer: $f (x) = 4 x^{3} - 2 x + 3$

Show working and answer

We integrate $f^{'} (x)$ : $f (x) = \int (12 x^{2} - 2) d x = 4 x^{3} - 2 x + C .$
We use $f (- 1) = 1$ : $4 (- 1)^{3} - 2 (- 1) + C = 1 \Rightarrow - 4 + 2 + C = 1 \Rightarrow C = 3.$

Answer: $f (x) = 4 x^{3} - 2 x + 3$

Exercise 68 BasicIB · P1 TZ0 November 2011

Let $f^{'} (x) = 3 x^{2} + 2$ . Given that $f (2) = 5$ , find $f (x)$ .

Show answer

Answer: $f (x) = x^{3} + 2 x - 7$

Show working and answer

We integrate $f^{'} (x)$ : $f (x) = \int (3 x^{2} + 2) d x = x^{3} + 2 x + C .$
We use $f (2) = 5$ : $8 + 4 + C = 5 \Rightarrow C = - 7.$

Answer: $f (x) = x^{3} + 2 x - 7$

Definite integrals and area

Exercise 69 BasicIB · P1 TZ1 May 2014

Let $f (x) = x^{2}$ . Calculate $\int_{1}^{2} (f (x))^{2} d x$ .

Show answer

Answer: $\frac{31}{5} = 6.2$

Show working and answer

We calculate $(f (x))^{2} = (x^{2})^{2} = x^{4}$ .
We integrate: $\int_{1}^{2} x^{4} d x = [\frac{x ^{5}}{5}]_{1}^{2} = \frac{32}{5} - \frac{1}{5} = \frac{31}{5} .$

Answer: $\frac{31}{5} = 6.2$

Exercise 70 BasicIB · P1 TZ2 May 2018

Let $f (x) = 6 x^{2} - 3 x$ . The following figure shows the graph of $f$ .

Graph of f(x)=6x²−3x between x=0 and x=2, dipping slightly below the x-axis near the origin and passing through (1; 3) and (2; 18)

a) Find $\int (6 x^{2} - 3 x) d x$ .

b) Find the area of the region bounded by the graph of $f$ , the $x$ -axis and the lines $x = 1$ and $x = 2$ .

Show answer

Answer: a) $2 x^{3} - \frac{3 x ^{2}}{2} + C$ ; b) Area $= 9.5$ square units

Show working and answer

We integrate term by term: $\int (6 x^{2} - 3 x) d x = 2 x^{3} - \frac{3 x ^{2}}{2} + C .$

On $1 \leq x \leq 2$ the function is positive ( $f (1) = 3$ , $f (2) = 18$ ), so the area is the integral of $f$ from $1$ to $2$ .
We evaluate: $\int_{1}^{2} (6 x^{2} - 3 x) d x = [2 x^{3} - \frac{3 x ^{2}}{2}]_{1}^{2} = (16 - 6) - (2 - \frac{3}{2}) = 10 - \frac{1}{2} = 9.5.$

Answer: a) $2 x^{3} - \frac{3 x ^{2}}{2} + C$ ; b) Area $= 9.5$ square units

Exercise 71 BasicIB · P1 TZ0 November 2012

Part of the graph of $f (x) = x - 4$ , for $x \geq 4$ , is shown below.

Graph of f(x)=√(x−4) from (4; 0) to (10; 2.45), with the region R under the curve between x=4 and x=10

The shaded region $R$ is enclosed by the graph of $f$ , the line $x = 10$ and the $x$ -axis. Calculate the area of $R$ .

Show answer

Answer: Area $(R) = 46 \approx 9.80$ square units

Show working and answer

The area is $A = \int_{4}^{10} x - 4 d x = \int_{4}^{10} (x - 4)^{1/2} d x .$
An antiderivative is $\frac{2}{3} (x - 4)^{3/2}$ . We evaluate: $A = \frac{2}{3} (6^{3/2} - 0) = \frac{2}{3} \cdot 66 = 46 \approx 9.80.$

Answer: Area $(R) = 46 \approx 9.80$ square units

Exercise 72 MediumIB · P1 TZ0 November 2014

The following figure shows the graph of $f (x) = \frac{x}{x ^{2} + 1}$ , for $0 \leq x \leq 4$ , and the line $x = 4$ .

Graph of f(x)=x/(x²+1) for 0 ≤ x ≤ 4, with the region R between the curve, the x-axis and the line x=4

Let $R$ be the region bounded by the graph of $f$ , the $x$ -axis and the line $x = 4$ . Find the area of $R$ .

Show answer

Answer: Area $(R) = \frac{1}{2} ln 17 \approx 1.42$ square units

Show working and answer

The area is the integral of $f$ from $0$ to $4$ : $A = \int_{0}^{4} \frac{x}{x ^{2} + 1} d x .$
The numerator is half the derivative of the denominator ( $(x^{2} + 1)^{'} = 2 x$ ), so an antiderivative is $\frac{1}{2} ln (x^{2} + 1)$ .
We evaluate: $A = [\frac{1}{2} ln (x^{2} + 1)]_{0}^{4} = \frac{1}{2} ln 17 - \frac{1}{2} ln 1 = \frac{1}{2} ln 17 \approx 1.42.$

Answer: Area $(R) = \frac{1}{2} ln 17 \approx 1.42$ square units

Exercise 73 MediumIB · P2 TZ2 May 2014

Let $f (x) = 5 - x^{2}$ . The following figure shows part of the graph of $f$ .

Graph of f(x)=5−x², a parabola with vertex (0; 5) crossing the x-axis at A and B on either side of the origin and continuing below the x-axis

The graph crosses the $x$ -axis at the points $A$ and $B$ .

a) Find the $x$ -coordinate of $A$ and of $B$ .

b) Find the area of the region bounded by the graph of $f$ and the $x$ -axis.

Show answer

Answer: a) $A$ : $x = - 5 \approx - 2.24$ ; $B$ : $x = 5 \approx 2.24$ ; b) Area $= \frac{20 5}{3} \approx 14.9$ square units

Show working and answer

The intersections with the $x$ -axis satisfy $5 - x^{2} = 0$ , that is, $x^{2} = 5$ .
$A$ has $x = - 5 \approx - 2.24$ and $B$ has $x = 5 \approx 2.24$ .

The region lies between the parabola and the $x$ -axis, from $- 5$ to $5$ : $Area = \int_{- 5}^{5} (5 - x^{2}) d x .$
Using the GDC (or by symmetry, $2 \int_{0}^{5}$ ): $2 [5 x - \frac{x ^{3}}{3}]_{0}^{5} = 2 (55 - \frac{5 5}{3}) = \frac{20 5}{3} \approx 14.9.$

Answer: a) $A$ : $x = - 5 \approx - 2.24$ ; $B$ : $x = 5 \approx 2.24$ ; b) Area $= \frac{20 5}{3} \approx 14.9$ square units

Exercise 74 MediumIB · P2 TZ2 May 2009

Let $f (x) = x (x - 5)^{2}$ , for $0 \leq x \leq 6$ . The following diagram shows the graph of $f$ .

Graph of f(x)=x(x−5)² for 0 ≤ x ≤ 6, with a local maximum near x=1.67 and the region R under the first arch, between x=0 and x=5

Let $R$ be the region enclosed by the $x$ -axis and the curve of $f$ . Calculate the area of $R$ .

Show answer

Answer: Area $(R) = \frac{625}{12} \approx 52.1$ square units

Show working and answer

As $f (x) \geq 0$ for $0 \leq x \leq 5$ , the area is the integral of $f$ from $0$ to $5$ .
Expanding: $x (x - 5)^{2} = x^{3} - 10 x^{2} + 25 x$ .
We integrate and evaluate: $A = [\frac{x ^{4}}{4} - \frac{10 x ^{3}}{3} + \frac{25 x ^{2}}{2}]_{0}^{5} = \frac{625}{4} - \frac{1250}{3} + \frac{625}{2} = \frac{1875 - 5000 + 3750}{12} = \frac{625}{12} \approx 52.1.$

Answer: Area $(R) = \frac{625}{12} \approx 52.1$ square units

Exercise 75 MediumIB · P2 TZ1 May 2018

Let $g (x) = - (x - 1)^{2} + 5$ .

a) Write down the coordinates of the vertex of the graph of $g$ .

Let $f (x) = x^{2}$ . The following diagram shows part of the graph of $f$ .

Graph of f(x)=x², a parabola with vertex at the origin opening upwards, with grid; its branches leave the drawing at the top

The graph of $g$ intersects the graph of $f$ at $x = - 1$ and $x = 2$ .

b) On the grid, draw the graph of $g$ for $- 2 \leq x \leq 4$ .

c) Find the area of the region enclosed by the graphs of $f$ and $g$ .

Show answer

Answer: a) $(1, 5)$ ; b) graph (figure); c) Area $= 9$ square units

Show working and answer

$g (x) = - (x - 1)^{2} + 5$ is in vertex form, so the vertex is $(1, 5)$ .

$g$ is a parabola opening downwards with vertex $(1, 5)$ ; it crosses the $y$ -axis at $(0, 4)$ and passes through $(- 1, 1)$ and $(2, 4)$ , where it meets $f$ .

Graphs of f(x)=x² and g(x)=−(x−1)²+5 for −2 ≤ x ≤ 4, with grid; they meet at (−1; 1) and (2; 4) and g has vertex V(1; 5)

Between $x = - 1$ and $x = 2$ the graph of $g$ lies above that of $f$ . Expanding $g (x) = - x^{2} + 2 x + 4$ , we get $g (x) - f (x) = - 2 x^{2} + 2 x + 4.$
The area is $A = \int_{- 1}^{2} (- 2 x^{2} + 2 x + 4) d x = [- \frac{2 x ^{3}}{3} + x^{2} + 4 x]_{- 1}^{2} = \frac{20}{3} - (- \frac{7}{3}) = 9.$

Answer: a) $(1, 5)$ ; b) graph (figure); c) Area $= 9$ square units

Exercise 76 MediumIB · P1 TZ2 May 2011

The following diagram shows part of the graph of the function $f (x) = 2 x^{2}$ .

Graph of f(x)=2x² near the origin, with the point (1; 2), the tangent T at that point and the shaded region R under the curve between the x-axis and T

The line $T$ is the tangent to the graph of $f$ at $x = 1$ .

a) Show that the equation of $T$ is $y = 4 x - 2$ .

b) Find the intersection of the $x$ -axis with $T$ .

c) The shaded region $R$ is enclosed by the graph of $f$ , the line $T$ and the $x$ -axis.

(i) Write an expression for the area of $R$ .

(ii) Calculate the area of $R$ .

Show answer

Answer: a) shown; b) $(\frac{1}{2}, 0)$ ; c) (i) $\int_{0}^{1} 2 x^{2} d x - \int_{1/2}^{1} (4 x - 2) d x$ ; (ii) $\frac{1}{6} \approx 0.167$ square units

Show working and answer

$f (1) = 2$ , so the point of tangency is $(1, 2)$ .
$f^{'} (x) = 4 x$ , so $f^{'} (1) = 4$ .
The tangent is $y - 2 = 4 (x - 1) \Rightarrow y = 4 x - 2.$

We set $y = 0$ : $4 x - 2 = 0$ , so $x = \frac{1}{2}$ . The intersection with the $x$ -axis is $(\frac{1}{2}, 0)$ .

(i) $R$ is the area under $f$ from $0$ to $1$ minus the area under $T$ from $\frac{1}{2}$ to $1$ : $Area (R) = \int_{0}^{1} 2 x^{2} d x - \int_{1/2}^{1} (4 x - 2) d x .$
(ii) We calculate each integral: $\int_{0}^{1} 2 x^{2} d x = \frac{2}{3}, \int_{1/2}^{1} (4 x - 2) d x = [2 x^{2} - 2 x]_{1/2}^{1} = 0 - (\frac{1}{2} - 1) = \frac{1}{2} .$
Hence $Area (R) = \frac{2}{3} - \frac{1}{2} = \frac{1}{6} \approx 0.167.$

Answer: a) shown; b) $(\frac{1}{2}, 0)$ ; c) (i) $\int_{0}^{1} 2 x^{2} d x - \int_{1/2}^{1} (4 x - 2) d x$ ; (ii) $\frac{1}{6} \approx 0.167$ square units

Exercise 77 MediumIB · P1 TZ0 November 2011

Let $f (x) = \frac{1}{4} x^{2} + 2$ . The line $L$ is the tangent to the curve of $f$ at $(4, 6)$ .

a) Find the equation of $L$ .

Let $g (x) = \frac{90}{3 x + 4}$ , for $2 \leq x \leq 12$ . The following diagram shows the graph of $g$ .

Graph of g(x)=90/(3x+4) for 2 ≤ x ≤ 12, decreasing and convex, from (2; 9) to (12; 2.25), with a very faint grid

b) Find the area of the region enclosed by the curve of $g$ , the $x$ -axis, and the lines $x = 2$ and $x = 12$ . Give your answer in the form $a ln b$ , where $a, b \in Z$ .

Show answer

Answer: a) $y = 2 x - 2$ ; b) Area $= 30 ln 4 \approx 41.6$ square units

Show working and answer

$f^{'} (x) = \frac{x}{2}$ , so $f^{'} (4) = 2$ .
The tangent at $(4, 6)$ is $y - 6 = 2 (x - 4) \Rightarrow y = 2 x - 2.$

The area is $A = \int_{2}^{12} \frac{90}{3 x + 4} d x .$
The derivative of $3 x + 4$ is $3$ , so an antiderivative is $30 ln (3 x + 4)$ (because $\frac{90}{3} = 30$ ).
We evaluate: $A = 30 [ln (3 x + 4)]_{2}^{12} = 30 (ln 40 - ln 10) = 30 ln \frac{40}{10} = 30 ln 4.$

Answer: a) $y = 2 x - 2$ ; b) Area $= 30 ln 4 \approx 41.6$ square units

Exercise 78 MediumIB · IB exam

Consider the area enclosed by the curve $y = 5 - \frac{x ^{3}}{25}$ and the positive $x$ and $y$ axes.

a) Sketch the curve, shading the required area.

b) Using the trapezium rule with $n = 5$ , approximate the value of the shaded area in part (a).

c) Determine the exact value of the shaded area.

d) Find the percentage error of the approximation in part (b).

Show answer

Answer: a) shaded area between the curve, the $x$ -axis and the $y$ -axis (from $x = 0$ to $x = 5$ ); b) $18.5$ ; c) $18.75$ ; d) $\approx 1.33 %$

Show working and answer

The curve falls from $(0, 5)$ and crosses the $x$ -axis at $x = 5$ (since $5 - \frac{125}{25} = 0$ ). The required area is the region between the $y$ -axis, the $x$ -axis and the curve, from $x = 0$ to $x = 5$ , which is shaded.

Graph of y=5−x³/25 for −3.5 ≤ x ≤ 6 and 0 ≤ y ≤ 7, with the region between the y-axis, the x-axis and the curve shaded up to x=5

With $n = 5$ trapezia on $[0, 5]$ , the width is $h = 1$ . Values: $f (0) = 5$ , $f (1) = 4.96$ , $f (2) = 4.68$ , $f (3) = 3.92$ , $f (4) = 2.44$ , $f (5) = 0$ .
Trapezium rule: $A \approx \frac{h}{2} [f (0) + 2 (f (1) + f (2) + f (3) + f (4)) + f (5)] = \frac{1}{2} [5 + 2 (16) + 0] = 18.5.$

We integrate: $A = \int_{0}^{5} (5 - \frac{x ^{3}}{25}) d x = [5 x - \frac{x ^{4}}{100}]_{0}^{5} = 25 - \frac{625}{100} = 18.75.$

The percentage error is $ε = \frac{∣18.5 - 18.75∣}{18.75} \cdot 100 = \frac{0.25}{18.75} \cdot 100 \approx 1.33 %.$

Answer: a) shaded area between the curve, the $x$ -axis and the $y$ -axis (from $x = 0$ to $x = 5$ ); b) $18.5$ ; c) $18.75$ ; d) $\approx 1.33 %$

Combined contextual problems

Exercise 79 MediumIB · P1 TZ0 November 2017

María owns a cheese factory. The quantity of cheese, $Q$ , in kilograms, that María sells in one week is given by $Q = 882 - 45 p$ , where $p$ is the price in euros (EUR) of one kilogram of cheese.

a) Write down how many kilograms of cheese María sells in one week if the price of one kilogram of cheese is 8 EUR.

María earns $(p - 6.80)$ EUR for each kilogram of cheese she sells.

b) Find how much money María earns in one week from selling cheese, if the price of one kilogram of cheese is 8 EUR.

To calculate her weekly profit $W$ , in EUR, María multiplies the quantity of cheese she sells by the money she earns from each kilogram.

c) Write an expression for $W$ in terms of $p$ .

d) Find the price, $p$ , that gives María the greatest weekly profit.

Show answer

Answer: a) $522$ kg; b) $626.40$ EUR; c) $W = - 45 p^{2} + 1188 p - 5997.6$ ; d) $p = 13.20$ EUR

Show working and answer

$Q = 882 - 45 \cdot 8 = 882 - 360 = 522$ kg.

She earns $8 - 6.80 = 1.20$ EUR per kilogram.
Weekly earnings: $522 \cdot 1.20 = 626.40$ EUR.

$W = Q \cdot (p - 6.80)$ , so $W = (882 - 45 p) (p - 6.80) = - 45 p^{2} + 1188 p - 5997.6.$

Differentiate: $\frac{d W}{d p} = - 90 p + 1188$ .
Set equal to zero: $- 90 p + 1188 = 0 \Rightarrow p = \frac{1188}{90} = 13.2.$
It is a maximum because $\frac{d ^{2} W}{d p ^{2}} = - 90 < 0$ .
The price that gives the greatest weekly profit is $13.20$ EUR per kilogram.

Answer: a) $522$ kg; b) $626.40$ EUR; c) $W = - 45 p^{2} + 1188 p - 5997.6$ ; d) $p = 13.20$ EUR

Exercise 80 MediumIB · P1 TZ2 May 2015

A construction company owns many rectangular plots of land, of different widths, along a road. The area, $A$ , of each plot is given by the function $A (x) = x (200 - x)$ , where $x$ represents the width of the plot in metres and $20 \leq x \leq 180$ .

a) Plot $S$ has a width of $20$ m. Write down the area of $S$ .

b) Plot $T$ has the same area as plot $S$ , but a different width. Find the width of $T$ .

When the width of the plot is $b$ metres, the plot has a maximum area.

c) (i) Write down the value of $b$ .

(ii) Write down the maximum area.

The range of $A (x)$ is $m \leq A (x) \leq n$ .

d) Hence write down the values of $m$ and $n$ .

Show answer

Answer: a) $3600 m^{2}$ ; b) $180$ m; c) (i) $b = 100$ m; (ii) $10000 m^{2}$ ; d) $m = 3600$ and $n = 10000$

Show working and answer

Substitute $x = 20$ into the area function: $A (20) = 20 (200 - 20) = 20 \cdot 180 = 3600 m^{2} .$

We look for another width with the same area: $x (200 - x) = 3600$ .
Rearranging the equation: $x^{2} - 200 x + 3600 = 0.$
Its solutions are $x = 20$ (plot $S$ ) and $x = 180$ .
Since the width of $T$ is different from $20$ m, the width of $T$ is $180$ m.

(i) Differentiate $A (x) = 200 x - x^{2}$ and set it equal to zero: $A^{'} (x) = 200 - 2 x = 0 \Rightarrow x = 100.$ Since $A^{'}$ changes from positive to negative at $x = 100$ , it is a maximum, and $b = 100$ m.
(ii) The maximum area is $A (100) = 100 \cdot (200 - 100) = 10000 m^{2} .$

On the interval $20 \leq x \leq 180$ the area takes its minimum value at the endpoints, $A (20) = A (180) = 3600$ , and its maximum value at $x = 100$ .
Therefore $m = 3600$ and $n = 10000$ .

Answer: a) $3600 m^{2}$ ; b) $180$ m; c) (i) $b = 100$ m; (ii) $10000 m^{2}$ ; d) $m = 3600$ and $n = 10000$

Exercise 81 MediumIB · P2 TZ1 May 2016

A distress flare is fired into the air from a ship at sea. The height, $h$ , in metres, of the flare above sea level is modelled by the quadratic function $h (t) = - 0.2 t^{2} + 16 t + 12$ , $t \geq 0$ , where $t$ is the time, in seconds, and $t = 0$ at the moment the flare is fired.

a) Write down the height from which the flare was fired.

b) Find the height of the flare 15 seconds after it was fired.

The flare falls into the sea $k$ seconds after it was fired.

c) Find the value of $k$ .

d) Find $h^{'} (t)$ .

e) (i) Show that the flare reached its maximum height 40 seconds after being fired.

(ii) Calculate the maximum height reached by the flare.

The nearest coastguard can see the flare when its height exceeds 40 metres above sea level.

f) Determine the total time for which the coastguard can see the flare.

Show answer

Answer: a) $12$ m; b) $207$ m; c) $k \approx 80.7$ s; d) $h^{'} (t) = - 0.4 t + 16$ ; e) (i) $h^{'} (t) = 0$ at $t = 40$ s (shown); (ii) $332$ m; f) $\approx 76.4$ s

Show working and answer

The flare is fired at $t = 0$ : $h (0) = 12 m .$

Substitute $t = 15$ : $h (15) = - 0.2 \cdot 225 + 16 \cdot 15 + 12 = - 45 + 240 + 12 = 207 m .$

The flare falls into the sea when $h (k) = 0$ , that is, $- 0.2 k^{2} + 16 k + 12 = 0$ .
Using the calculator (or the quadratic formula) we obtain two solutions, $k \approx - 0.74$ and $k \approx 80.7$ . The negative one makes no sense in the problem.
Therefore $k = 40 + 1660 \approx 80.7 s .$

Differentiate: $h^{'} (t) = - 0.4 t + 16.$

(i) At the maximum, $h^{'} (t) = 0$ : $- 0.4 t + 16 = 0 \Rightarrow t = \frac{16}{0.4} = 40 s .$ Since $h^{'}$ is positive before $t = 40$ and negative after, there is a maximum: the flare reaches its maximum height 40 s after being fired.
(ii) The maximum height is $h (40) = - 0.2 \cdot 1600 + 16 \cdot 40 + 12 = - 320 + 640 + 12 = 332 m .$

The coastguard can see the flare when $h (t) > 40$ . We solve $h (t) = 40$ : $- 0.2 t^{2} + 16 t + 12 = 40 \Rightarrow t^{2} - 80 t + 140 = 0.$
The solutions are $t = 40 \pm 1460$ , that is, $t_{1} \approx 1.79$ s and $t_{2} \approx 78.21$ s.
The total visibility time is $t_{2} - t_{1} = 21460 \approx 76.4 s .$

Answer: a) $12$ m; b) $207$ m; c) $k \approx 80.7$ s; d) $h^{'} (t) = - 0.4 t + 16$ ; e) (i) $h^{'} (t) = 0$ at $t = 40$ s (shown); (ii) $332$ m; f) $\approx 76.4$ s

Exercise 82 MediumIB · P2 TZ2 May 2013

The diagram below shows an aerial view of a cycle path. The shape of this cycle path can be modelled by the quadratic function $y = - \frac{x ^{2}}{10} + \frac{27}{2} x$ , where $x \geq 0$ , $y \geq 0$ .

$(x, y)$ are the coordinates of a point located $x$ metres east and $y$ metres north of $O$ , where $O$ is the origin $(0, 0)$ . $B$ is a point on the cycle path with coordinates $(100, 350)$ .

Aerial view of the cycle path on axes with x from 0 to 120 m and y from 0 to 500 m: an arc starting at the origin O, rising to a maximum near (67.5; 456) and falling to (120; 180); the point B(100; 350) lies on the curve, and on the right there is an arrow labelled N (north)

a) The coordinates of the point $A$ are $(75, 450)$ .

Determine whether or not the point $A$ lies on the cycle path. Give a reasoned answer.

b) Find the derivative of $y = - \frac{x ^{2}}{10} + \frac{27}{2} x$ .

c) Use your answer to part (b) to determine whether $A (75, 450)$ is the point on the cycle path between $O$ and $B$ that is furthest north. Give a reasoned answer.

d) (i) Write down the coordinates of the midpoint of the line segment $O B$ .

(ii) Find the gradient of the line segment $O B$ .

Scott leaves on his bike from a point $C (0, 150)$ . He heads towards the cycle path, along a straight road that is parallel to the line segment $O B$ .

e) Find the equation of the road Scott is travelling on. Express your answer in the form $a x + b y = c$ , where $a$ , $b$ and $c \in R$ .

f) Use your GDC to find the coordinates of the point where Scott first crosses the cycle path.

Show answer

Answer: a) yes, it lies on the cycle path; b) $\frac{d y}{d x} = - \frac{x}{5} + \frac{27}{2}$ ; c) it is not the furthest north point (the maximum is at $x = 67.5$ ); d) (i) $(50, 175)$ ; (ii) $3.5$ ; e) $7 x - 2 y = - 300$ ; f) $(18.4; 214)$

Show working and answer

We substitute $x = 75$ into the equation: $y = - \frac{7 5 ^{2}}{10} + \frac{27}{2} \cdot 75 = - 562.5 + 1012.5 = 450.$
This matches the $y$ -coordinate of $A$ , so $A$ does lie on the cycle path.

b) $\frac{d y}{d x} = - \frac{x}{5} + \frac{27}{2} .$

At $A$ : $\frac{d y}{d x} = - \frac{75}{5} + 13.5 = - 1.5 \neq = 0$ . The curve does not have a horizontal tangent at $A$ , and it is already decreasing there.
The furthest north point satisfies $\frac{d y}{d x} = 0$ : $- \frac{x}{5} + \frac{27}{2} = 0 \Rightarrow x = 67.5,$ and there $y = - \frac{67. 5 ^{2}}{10} + \frac{27}{2} \cdot 67.5 \approx 455.6 > 450$ .
Since $67.5$ lies between $0$ and $100$ , $A$ is not the furthest north point between $O$ and $B$ .

(i) Midpoint: $(\frac{0 + 100}{2}, \frac{0 + 350}{2}) = (50, 175)$ .
(ii) Gradient: $m_{O B} = \frac{350 - 0}{100 - 0} = 3.5$ .

The road is parallel to $O B$ , so it has gradient $3.5$ and passes through $C (0, 150)$ : $y = 3.5 x + 150.$
Multiplying by $2$ and rearranging: $7 x - 2 y = - 300$ .

Using the GDC, find the points of intersection of $y = - \frac{x ^{2}}{10} + \frac{27}{2} x$ and $y = 3.5 x + 150$ (this is equivalent to $x^{2} - 100 x + 1500 = 0$ ).
The $x$ -coordinates are $x \approx 18.4$ and $x \approx 81.6$ . Scott starts from $x = 0$ , so he first crosses at $x \approx 18.4$ .
Then $y = 3.5 \cdot 18.377 + 150 \approx 214$ .

Exercise 83 HardIB · P2 TZ0 November 2018

Haruka has an eco-friendly bag in the shape of a cuboid that is 12 cm wide, 36 cm long and 9 cm high. The bag is made from five rectangular pieces of fabric and is open at the top.

Fabric bag in the shape of a cuboid, open at the top, with two arched handles; dimensions of 36 cm length, 12 cm width and 9 cm height

a) Calculate the area of fabric, in $cm^{2}$ , needed to make Haruka's bag.

b) Calculate the volume of the bag, in $cm^{3}$ .

Nanako decides to make her own eco-friendly bag in the shape of a cuboid, so that its surface area is minimised.

Nanako's bag has a width of $x$ cm, a length that is three times its width and a height equal to $y$ cm.

Fabric bag in the shape of a cuboid, open at the top, nearly cubic, with two arched handles; dimensions x (width), 3x (length) and y (height)

The volume of Nanako's bag is equal to $3888 cm^{3}$ .

c) Use this value to write down, and simplify, an equation in $x$ and $y$ for the volume of Nanako's bag.

d) Write down and simplify an expression in $x$ and $y$ for the area of fabric, $A$ , used to make Nanako's bag.

e) Use your answers to parts (c) and (d) to show that $A = 3 x^{2} + \frac{10368}{x}$ .

f) Find $\frac{d A}{d x}$ .

g) Use your answer to part (f) to show that the width of Nanako's bag is equal to $12$ cm.

The fabric used to make Nanako's bag costs 4 Japanese yen (JPY) per $cm^{2}$ .

h) Find the cost of the fabric used to make Nanako's bag.

Show answer

Answer: a) $1296 cm^{2}$ ; b) $3888 cm^{3}$ ; c) $x^{2} y = 1296$ ; d) $A = 3 x^{2} + 8 x y$ ; e) shown; f) $\frac{d A}{d x} = 6 x - \frac{10 368}{x ^{2}}$ ; g) $x = 12$ cm (shown); h) $5184$ JPY

Show working and answer

The bag has five pieces: the base of $12 \times 36 = 432 cm^{2}$ , two sides of $36 \times 9 = 324 cm^{2}$ and two sides of $12 \times 9 = 108 cm^{2}$ .
Area of fabric: $A = 432 + 2 (324) + 2 (108) = 1296 cm^{2} .$

$V = 12 \cdot 36 \cdot 9 = 3888 cm^{3}$ .

With width $x$ , length $3 x$ and height $y$ : $V = x \cdot 3 x \cdot y = 3 x^{2} y = 3888$ .
Simplifying: $x^{2} y = 1296$ .

Base: $x \cdot 3 x = 3 x^{2}$ . Two faces of $3 x \times y$ : $6 x y$ . Two faces of $x \times y$ : $2 x y$ .
$A = 3 x^{2} + 8 x y$ .

From (c): $y = \frac{1296}{x ^{2}}$ .
Substitute into (d): $A = 3 x^{2} + 8 x \cdot \frac{1296}{x ^{2}} = 3 x^{2} + \frac{10 368}{x} .$

$\frac{d A}{d x} = 6 x - \frac{10 368}{x ^{2}}$ .

At the minimum, $\frac{d A}{d x} = 0$ : $6 x = \frac{10 368}{x ^{2}} \Rightarrow x^{3} = 1728 \Rightarrow x = 12 cm .$

With $x = 12$ : $A = 3 \cdot 144 + \frac{10 368}{12} = 432 + 864 = 1296 cm^{2}$ .
Cost: $4 \cdot 1296 = 5184$ JPY.

Answer: a) $1296 cm^{2}$ ; b) $3888 cm^{3}$ ; c) $x^{2} y = 1296$ ; d) $A = 3 x^{2} + 8 x y$ ; e) shown; f) $\frac{d A}{d x} = 6 x - \frac{10 368}{x ^{2}}$ ; g) $x = 12$ cm (shown); h) $5184$ JPY

Exercise 84 HardIB · P2 TZ2 May 2017

Violeta plans to grow flowers on a rectangular piece of land. She places a fence to mark the perimeter of the plot and uses 200 metres of fencing for this.

The plot of land is $x$ metres long.

Vertical rectangle drawn with a thick border representing the plot of land, with a vertical dimension x on its left side

a) Show that the width of the plot, in metres, is given by $100 - x$ .

b) Write down the area of the plot in terms of $x$ .

Violeta places the fencing so as to maximise the area of the plot.

c) Find the value of $x$ that maximises the area of the plot.

From selling the flowers, Violeta earns 2 Bulgarian lev (BGN) for each square metre of land.

d) Show that Violeta earns 5000 BGN from selling the flowers she has grown on this plot.

Violeta wants to invest her 5000 BGN.

e) A bank offers her a nominal annual interest rate of 4%, compounded semi-annually.

(i) Find the amount of money Violeta would have after 6 years. Give your answer correct to two decimal places.

(ii) Find how long it would take for the interest received to be 2000 BGN.

Another bank offers her an interest rate of $r %$ compounded annually, which would allow her to double her money in 12 years.

f) Find the smallest possible value of $r$ .

Show answer

Answer: a) $w = 100 - x$ (shown); b) $A (x) = 100 x - x^{2}$ ; c) $x = 50$ m; d) $2500 m^{2} \cdot 2 BGN/m^{2} = 5000$ BGN (shown); e) (i) $6341.21$ BGN; (ii) $8.5$ years (17 half-years); f) $r \approx 5.95 %$

Show working and answer

Let $w$ be the width. The perimeter is $2 x + 2 w = 200$ .
Solving for $w$ : $w = 100 - x .$

$A (x) = x (100 - x) = 100 x - x^{2}$ .

Differentiate: $A^{'} (x) = 100 - 2 x$ .
$A^{'} (x) = 0 \Rightarrow x = 50$ m. It is a maximum because $A^{''} (x) = - 2 < 0$ .

With $x = 50$ , the width is $100 - 50 = 50$ m and the maximum area is $50 \cdot 50 = 2500 m^{2}$ .
Earnings: $2 \cdot 2500 = 5000$ BGN.

(i) Interest of $4 %$ per year compounded semi-annually: each half-year $2 %$ is applied. In $6$ years there are $12$ half-years: $5000 (1 + \frac{4}{200})^{12} = 5000 (1.02)^{12} \approx 6341.21 BGN .$
(ii) With interest of $2000$ BGN the amount would be $7000$ BGN. Using the GDC we solve $5000 (1.02)^{n} = 7000 \Rightarrow n \approx 16.99 half-years .$
This is $17$ half-years, that is, $8.5$ years.

With annual interest $r %$ over $12$ years: $5000 (1 + \frac{r}{100})^{12} = 10000 \Rightarrow 1 + \frac{r}{100} = 2^{1/12} .$
Therefore $r = 100 (2^{1/12} - 1) \approx 5.946$ , and the smallest possible value is $r \approx 5.95 %$ .

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IB exam questions for topic 7

Derivatives and differentiation rules

Tangent and normal lines

Increasing and decreasing, maxima and minima

Optimisation

Antiderivatives and indefinite integrals

Definite integrals and area

Combined contextual problems

Other problems

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