IB Analysis and Approaches HL · Functions · Section 7 of 15

Combining functions and composition

3 worked examples · By Álvaro López

Key idea

Functions can be combined like numbers: added, multiplied or divided, and also chained, so that the output of one becomes the input of another. This last operation is called composition, and it is the basis of the inverse function and of the chain rule in topic 10.

Sum, difference, product and quotient

Given two functions $f$ and $g$ , new ones are defined by operating on their values at each $x$ :

The domain of the sum, the difference and the product is the set of $x$ that are in the domain of $f$ and in the domain of $g$ at the same time. For the quotient, you must also remove the $x$ for which $g (x) = 0$ .

Composition of functions

Read it as " $f$ composed with $g$ ", and apply it from the inside out: in $f (g (x))$ the inner function, $g$ , acts first. To write the formula, replace each $x$ in $f$ by the full expression of $g (x)$ .

Composition is not commutative: in general, $f \circ g \neq = g \circ f$ . With $f (x) = x + 1$ and $g (x) = x^{2}$ you get $(f \circ g) (x) = x^{2} + 1$ , but $(g \circ f) (x) = (x + 1)^{2}$ .

Domain of the composition

It is important to find the domain before simplifying the formula. In the example below, one of the compositions simplifies to an expression that seems to accept more values than it really does.

Worked examples

Example 1 Basic

Operating on and composing two functions

Let $f (x) = 2 x + 1$ and $g (x) = x^{2} - 3$ . Find:

a) $(f + g) (x)$ .

b) $(f g) (x)$ expanded.

c) $(f \circ g) (x)$ .

d) $(g \circ f) (x)$ expanded.

Solution

a) $(f + g) (x) = 2 x + 1 + x^{2} - 3 = x^{2} + 2 x - 2$ .

b) $(f g) (x) = (2 x + 1) (x^{2} - 3) = 2 x^{3} - 6 x + x^{2} - 3 = 2 x^{3} + x^{2} - 6 x - 3$ .

c) Substitute $g (x)$ into $f$ : $(f \circ g) (x) = 2 (x^{2} - 3) + 1 = 2 x^{2} - 5$ .

d) Substitute $f (x)$ into $g$ : $(g \circ f) (x) = (2 x + 1)^{2} - 3 = 4 x^{2} + 4 x - 2$ .

Note that $f \circ g \neq = g \circ f$ .

Example 2 Medium

Domains of a composition and of a quotient

Let $f (x) = x$ and $g (x) = x^{2} - 9$ .

a) Find the formula of $(f \circ g) (x)$ .

b) Find the domain of $f \circ g$ .

c) Find the formula of $(g \circ f) (x)$ in its simplest form.

d) Find the domain of $g \circ f$ .

e) Find the domain of $\frac{f}{g}$ .

Solution

a) $(f \circ g) (x) = x^{2} - 9$ .

b) $g$ is defined for all $x$ . You need $g (x)$ to be in the domain of $f$ , that is, $x^{2} - 9 \geq 0$ : $x \leq - 3$ or $x \geq 3$ . The domain is $] - \infty, - 3] \cup [3, + \infty [$ .

c) $(g \circ f) (x) = (x)^{2} - 9 = x - 9$ .

d) Although $x - 9$ is defined for all $x$ , the domain of the composite requires $x$ to be in the domain of $f$ , which is $x \geq 0$ . Then $f (x) = x \geq 0$ is in the domain of $g$ (all reals). Domain: $[0, + \infty [$ .

e) You need $x \geq 0$ (domain of $f$ ) and $g (x) \neq = 0$ , that is, $x \neq = 3$ (and $x \neq = - 3$ , which is already excluded). Domain: $[0, 3 [\cup] 3, + \infty [$ .

Example 3 Hard

Finding a function from its composite

Let $f (x) = 3 x - 2$ . The function $g$ satisfies $(f \circ g) (x) = 6 x^{2} + 1$ .

a) Find $g (x)$ .

b) Find $(g \circ f) (x)$ expanded.

c) Solve the equation $(g \circ f) (x) = 3$ .

d) Find $(f \circ f) (x)$ .

Solution

a) By definition, $f (g (x)) = 3 g (x) - 2$ . Equate it with $6 x^{2} + 1$ : $3 g (x) - 2 = 6 x^{2} + 1 ⟹ g (x) = \frac{6 x ^{2} + 3}{3} = 2 x^{2} + 1$ Check: $f (g (x)) = 3 (2 x^{2} + 1) - 2 = 6 x^{2} + 1$ .

b) $(g \circ f) (x) = g (3 x - 2) = 2 (3 x - 2)^{2} + 1 = 2 (9 x^{2} - 12 x + 4) + 1 = 18 x^{2} - 24 x + 9$ .

c) $18 x^{2} - 24 x + 9 = 3 \Rightarrow 18 x^{2} - 24 x + 6 = 0 \Rightarrow 3 x^{2} - 4 x + 1 = 0 \Rightarrow (3 x - 1) (x - 1) = 0$ . The solutions are $x = \frac{1}{3}$ and $x = 1$ .

d) $(f \circ f) (x) = f (3 x - 2) = 3 (3 x - 2) - 2 = 9 x - 8$ .

In the exam: make the order clear. Write " $f (g (x)) = \dots$ " and substitute with brackets. When you have to find an unknown function, check the result by substituting: it is an easy mark to secure.

Common mistakes

Summary

$(f \pm g) (x) = f (x) \pm g (x)$ , $(f g) (x) = f (x) g (x)$ , $(\frac{f}{g}) (x) = \frac{f ( x )}{g ( x )}$ with $g (x) \neq = 0$ .
$(f \circ g) (x) = f (g (x))$ : the inner function acts first. In general, $f \circ g \neq = g \circ f$ .
Domain of $f \circ g$ : the $x$ -values in the domain of $g$ for which $g (x)$ is in the domain of $f$ .
The domain is calculated before simplifying the formula.
To find $g$ from $f \circ g$ , equate $f (g (x))$ with the given expression and solve for $g (x)$ .

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