IB Analysis and Approaches HL · Functions · Section 8 of 15

Inverse function

3 worked examples · By Álvaro López

Key idea

The inverse function undoes what the function does: if $f$ turns $x$ into $y$ , the inverse turns $y$ back into $x$ . It exists only if each output comes from a single input, and its graph is the reflection of that of $f$ in the line $y = x$ . In the exam you are asked for its formula, its domain and the point where it meets $f$ .

What the inverse is and when it exists

A function with this property is called injective (or one-to-one). With the horizontal line test you can check it on the graph: if any horizontal line cuts the curve at more than one point, that output comes from two inputs and $f$ has no inverse. The parabola $f (x) = x^{2}$ has no inverse on all of $R$ , because $f (2) = f (- 2) = 4$ .

Be careful with the notation: $f^{- 1} (x)$ is not $\frac{1}{f ( x )}$ . The $- 1$ denotes the inverse function, not a power.

Domain and range of the inverse

As $f^{- 1}$ goes the opposite way, it swaps the roles of $x$ and $y$ :

This makes it easy to choose the domain of $f^{- 1}$ without calculations: you only need to know the values between which $f$ moves.

How to calculate the inverse

It is the same process as "solving for $x$ in terms of $y$ " and then changing the letters so that the function uses the usual variable. To check, calculate $f (f^{- 1} (x))$ , which must give $x$ .

Restricting the domain

If $f$ is not injective, its domain can be restricted to an interval where it is injective, and then that piece has an inverse. With $f (x) = x^{2}$ :

with $x \geq 0$ , the inverse is $f^{- 1} (x) = x$ ;
with $x \leq 0$ , the inverse is $f^{- 1} (x) = - x$ .

The choice of piece is what determines which of the two roots is the correct one.

Graph of the inverse

If the point $(a, b)$ is on the graph of $f$ , then $f (a) = b$ and $f^{- 1} (b) = a$ , that is, $(b, a)$ is on the graph of $f^{- 1}$ . By swapping the coordinates, each point is reflected in the line $y = x$ .

Graphs of f(x)=x^2 with x greater than or equal to 0, of its inverse square root of x and of the line y=x: they are symmetric about y=x

For an increasing function, the points where $f$ and $f^{- 1}$ meet lie on the line $y = x$ , so they are found by solving $f (x) = x$ .

Worked examples

Example 1 Basic

The inverse of a linear function

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

a) Find $f^{- 1} (x)$ .

b) Calculate $f^{- 1} (5)$ .

Solution

a) Write $y = \frac{2 x - 1}{3}$ and swap the letters: $x = \frac{2 y - 1}{3}$ . Solve for $y$ : $3 x = 2 y - 1 \Rightarrow y = \frac{3 x + 1}{2}$ . Therefore $f^{- 1} (x) = \frac{3 x + 1}{2}$

b) $f^{- 1} (5) = \frac{3 \cdot 5 + 1}{2} = 8$ . Check: $f (8) = \frac{16 - 1}{3} = 5$ .

Example 2 Medium

Inverse of a fraction

Let $f (x) = \frac{x + 1}{x - 2}$ , with $x \neq = 2$ .

a) Find $f^{- 1} (x)$ .

b) Find the domain of $f^{- 1}$ .

c) Find $f^{- 1} (4)$ .

Solution

a) Write $y = \frac{x + 1}{x - 2}$ and swap the letters: $x = \frac{y + 1}{y - 2}$ . Solve for $y$ : $x (y - 2) = y + 1 ⟹ x y - 2 x = y + 1 ⟹ y (x - 1) = 2 x + 1 ⟹ y = \frac{2 x + 1}{x - 1}$ Therefore $f^{- 1} (x) = \frac{2 x + 1}{x - 1}$ .

b) The domain of $f^{- 1}$ is the range of $f$ . The formula for $f^{- 1}$ fails at $x = 1$ , and the value $1$ is never an output of $f$ (if $\frac{x + 1}{x - 2} = 1$ then $x + 1 = x - 2$ , which is impossible). The domain is $R ∖ {1}$ .

c) $f^{- 1} (4) = \frac{2 \cdot 4 + 1}{4 - 1} = 3$ . Check: $f (3) = \frac{4}{1} = 4$ .

Example 3 Hard

A parabola with a restricted domain

Let $f (x) = (x - 2)^{2} + 1$ , defined for $x \geq 2$ .

a) Find the range of $f$ .

b) Find $f^{- 1} (x)$ and its domain.

c) Find the value of $x$ for which $f (x) = f^{- 1} (x)$ .

d) Explain why the restriction $x \geq 2$ is needed.

Solution

a) The vertex is $(2, 1)$ and $f$ is increasing for $x \geq 2$ , so the range is $[1, + \infty [$ .

b) Write $y = (x - 2)^{2} + 1$ and swap the letters: $x = (y - 2)^{2} + 1$ . Solving: $(y - 2)^{2} = x - 1$ , so $y - 2 = x - 1$ (the positive root is taken because the original branch satisfies $y \geq 2$ ). Then $f^{- 1} (x) = 2 + x - 1$ Its domain is the range of $f$ : $[1, + \infty [$ (and its range is $[2, + \infty [$ , the domain of $f$ ).

c) As $f$ is increasing, $f$ and $f^{- 1}$ meet on the line $y = x$ , so solve $f (x) = x$ : $(x - 2)^{2} + 1 = x ⟹ x^{2} - 5 x + 5 = 0 ⟹ x = \frac{5 \pm 5}{2}$ Only $x = \frac{5 + 5}{2} \approx 3.62$ satisfies $x \geq 2$ (the other is $\approx 1.38$ ).

d) Without the restriction, $f$ is not injective: $f (0) = f (4) = 5$ , and a horizontal line cuts the parabola twice. With $x \geq 2$ only the right-hand branch remains, on which each output comes from a single input.

In the exam: when finding the inverse of a quadratic, choose the sign of the root using the original domain, and do not forget the domain of the inverse. For $f (x) = f^{- 1} (x)$ it is convenient to use $f (x) = x$ , but check that the solution is in the domain.

Common mistakes

Summary

$f^{- 1}$ exists if $f$ is injective (horizontal line test); otherwise the domain is restricted.
$f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ ; $f^{- 1} (b) = a$ if and only if $f (a) = b$ .
Domain of $f^{- 1}$ = range of $f$ ; range of $f^{- 1}$ = domain of $f$ .
Calculation: $y = f (x)$ , swap $x$ and $y$ , solve for $y$ , write the domain.
The graph of $f^{- 1}$ is that of $f$ reflected in $y = x$ .
$f^{- 1} (x)$ is not $\frac{1}{f ( x )}$ .

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