Composition of Functions

Sometimes the output of one function becomes the input of another. The distance you drive depends on time, the fuel you burn depends on distance, and the cost depends on the fuel. Chaining functions like this is called composition, and it gives you a single function that goes straight from the first input to the final output.

Key ideas

What f(g(x)) means

The composition of $f$ and $g$ is

$$(f \circ g)(x) = f\big(g(x)\big)$$

Read it from the inside out: first apply $g$ to $x$, then apply $f$ to the result. Picture two function machines in a row. The number $x$ goes into machine $g$, and whatever comes out goes straight into machine $f$.

$$x \;\xrightarrow{\;g\;}\; g(x) \;\xrightarrow{\;f\;}\; f\big(g(x)\big)$$

For example, if $g(x) = x + 1$ and $f(x) = x^2$, then $f(g(3)) = f(4) = 16$.

Order matters

In general, $f(g(x)) \ne g(f(x))$. With the same functions, $g(f(3)) = g(9) = 10$, not $16$. “Add $1$, then square” is a different process from “square, then add $1$”. Occasionally the two orders do agree (for example, a function and its inverse), but you can never assume it.

Three ways to compose

• From tables: find $g(x)$ in the $g$ table, then look up that number in the $f$ table.
• From graphs: read $g(x)$ from the graph of $g$ (go up or down from $x$ to the graph). Then use that output as the new input on the graph of $f$.
• From equations: substitute the whole expression for $g(x)$ in place of every $x$ in $f(x)$. Use brackets.

Reading compositions from graphs: here $f(x) = x^2 - 2$ and $g(x) = -x + 3$.

To find $f(g(1))$ from the graphs above: the line gives $g(1) = 2$, and then the parabola gives $f(2) = 2$. So $f(g(1)) = 2$. In the other order, $f(1) = -1$, and then $g(-1) = 4$, so $g(f(1)) = 4$.

Domain and range of a composition

$x$ is in the domain of $f \circ g$ when both of these are true:

1. $x$ is in the domain of $g$ (so $g(x)$ exists), and
2. $g(x)$ is in the domain of $f$ (so $f$ can accept it).

Find the domain from these conditions, not just from the simplified formula, which can hide a restriction (Example 3 shows how). For the range, think about which values of $g(x)$ actually get fed into $f$, and what $f$ does to them.

Decomposing a function

Going backwards, you can often split a complicated function into an “inside” and an “outside” function. For $h(x) = (5x - 2)^3$, the inside is $g(x) = 5x - 2$ and the outside is $f(x) = x^3$, so $h(x) = f(g(x))$. Decomposing is useful for seeing how a function was built, and you’ll use it constantly in calculus.

Worked examples

Example 1: Composing from tables

Use the tables to evaluate $f(g(0))$, $g(f(0))$, $f(g(3))$, $g(f(3))$ and $f(f(4))$.

$x$	$0$	$1$	$2$	$3$	$4$
$f(x)$	$3$	$4$	$0$	$2$	$1$
$g(x)$	$1$	$3$	$4$	$0$	$2$

Solution. Work from the inside out each time.

• $f(g(0))$: $g(0) = 1$, then $f(1) = 4$. So $f(g(0)) = 4$.
• $g(f(0))$: $f(0) = 3$, then $g(3) = 0$. So $g(f(0)) = 0$.
• $f(g(3))$: $g(3) = 0$, then $f(0) = 3$. So $f(g(3)) = 3$.
• $g(f(3))$: $f(3) = 2$, then $g(2) = 4$. So $g(f(3)) = 4$.
• $f(f(4))$: $f(4) = 1$, then $f(1) = 4$. So $f(f(4)) = 4$.

The first two show that changing the order changes the answer: $f(g(0)) = 4$ but $g(f(0)) = 0$.

Example 2: Composing equations

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

Solution. For $f(g(x))$, replace the $x$ in $f$ with the whole expression $(3x - 2)$:

$$\begin{aligned} f\big(g(x)\big) &= (3x - 2)^2 + 1 \\ &= 9x^2 - 12x + 4 + 1 \\ &= 9x^2 - 12x + 5 \end{aligned}$$

For $g(f(x))$, replace the $x$ in $g$ with $(x^2 + 1)$:

$$g\big(f(x)\big) = 3(x^2 + 1) - 2 = 3x^2 + 1$$

The results are different functions.

Check with $x = 1$: $g(1) = 1$ and $f(1) = 2$, and the formula gives $9 - 12 + 5 = 2$. ✓ Also $f(1) = 2$ and $g(2) = 4$, and the formula gives $3 + 1 = 4$. ✓

Example 3: Domain and range

Let $f(x) = \sqrt{x}$ and $g(x) = 4 - x^2$. Find $f(g(x))$ and $g(f(x))$, and state the domain and range of each.

Solution.

$f(g(x))$.

$$f\big(g(x)\big) = \sqrt{4 - x^2}$$

$g$ accepts every real number, but $f$ can only take inputs that are $\ge 0$. So we need $4 - x^2 \ge 0$, which means $x^2 \le 4$, or $-2 \le x \le 2$.

For these $x$-values, $4 - x^2$ goes from $0$ (at $x = \pm 2$) up to $4$ (at $x = 0$). Taking square roots gives values from $0$ to $2$.

$$\text{Domain: } \{x \in \mathbb{R} \mid -2 \le x \le 2\} \qquad \text{Range: } \{y \in \mathbb{R} \mid 0 \le y \le 2\}$$

(The graph is the top half of a circle with radius $2$.)

$g(f(x))$.

$$g\big(f(x)\big) = 4 - \left(\sqrt{x}\right)^2 = 4 - x$$

It looks like a line, but the first step is $\sqrt{x}$, which needs $x \ge 0$. So the domain is $\{x \in \mathbb{R} \mid x \ge 0\}$, and the graph is only the part of $y = 4 - x$ starting at $(0, 4)$. As $x$ grows from $0$, $4 - x$ falls from $4$, so the range is $\{y \in \mathbb{R} \mid y \le 4\}$.

Example 4: Fuel cost as a function of time

A delivery van drives on the highway at $100$ km/h, so after $t$ hours it has gone $d(t) = 100t$ km. It uses $7.5$ L of fuel per $100$ km, so $F(d) = 0.075d$ litres for $d$ kilometres. Fuel costs $1.60 per litre, so $C(F) = 1.60F$ dollars.

• (a) Find $C(F(d(t)))$ and explain what it means.
• (b) How much does the fuel cost for a $4.5$-hour drive?

Solution.

(a) Work from the inside out:

$$\begin{aligned} F\big(d(t)\big) &= 0.075(100t) = 7.5t \\ C\big(F(d(t))\big) &= 1.60(7.5t) = 12t \end{aligned}$$

$C(F(d(t))) = 12t$ is the fuel cost, in dollars, of driving for $t$ hours: the van burns $12 of fuel per hour. Composition skipped the middle steps and went straight from time to cost.

(b) $12(4.5) = 54$, so the fuel costs $54.

Check step by step: $4.5$ h gives $450$ km, which uses $0.075(450) = 33.75$ L, which costs $1.60(33.75) = 54.00$, or $54.00. ✓

Common mistakes

Working from the outside in. $f(g(x))$ means $g$ first, then $f$, even though $f$ is written first. Start with the innermost brackets.

Multiplying instead of composing. $f(g(x))$ is not $f(x) \cdot g(x)$. For $f(x) = x^2 + 1$ and $g(x) = 3x - 2$, the product is $3x^3 - 2x^2 + 3x - 2$, but the composition is $9x^2 - 12x + 5$.

Forgetting brackets when substituting. $f(x) = x^2 + 1$ with $x = 3x - 2$ gives $(3x - 2)^2 + 1$, not $3x - 2^2 + 1$ or $3x^2 - 2 + 1$.

Reading the domain from the simplified formula. In Example 3, $g(f(x))$ simplifies to $4 - x$, but $x = -5$ isn’t allowed, because $f(-5) = \sqrt{-5}$ doesn’t exist. Check the inside function first.

Reusing the x-value on a graph. After you find $g(1) = 2$, the next input is $2$, not $1$. Move to $x = 2$ on the graph of $f$.

Assuming f(g(x)) = g(f(x)). It’s usually false. Example 1 and Example 2 both show different answers in the two orders.

Practice

1. (Warm-up) Use the tables to find $f(g(1))$, $g(f(1))$ and $g(g(2))$.

$x$	$1$	$2$	$3$	$4$
$f(x)$	$3$	$1$	$4$	$2$
$g(x)$	$2$	$4$	$1$	$3$

Solution

$f(g(1))$: $g(1) = 2$, then $f(2) = 1$. So $f(g(1)) = 1$.

$g(f(1))$: $f(1) = 3$, then $g(3) = 1$. So $g(f(1)) = 1$.

$g(g(2))$: $g(2) = 4$, then $g(4) = 3$. So $g(g(2)) = 3$.

2. (Warm-up) Let $f(x) = 2x + 1$ and $g(x) = x^2$. Find $f(g(3))$ and $g(f(3))$.

Solution

$f(g(3))$: $g(3) = 9$, then $f(9) = 19$.

$g(f(3))$: $f(3) = 7$, then $g(7) = 49$.

3. (Warm-up) Use the graphs of $f$ and $g$ in Key ideas to find $g(f(0))$, $f(g(2))$ and $f(f(1))$.

Solution

$g(f(0))$: the parabola gives $f(0) = -2$, then the line gives $g(-2) = 5$. So $g(f(0)) = 5$.

$f(g(2))$: the line gives $g(2) = 1$, then the parabola gives $f(1) = -1$. So $f(g(2)) = -1$.

$f(f(1))$: $f(1) = -1$, then $f(-1) = -1$. So $f(f(1)) = -1$.

4. (Core) Let $f(x) = x^2 - 3x$ and $g(x) = x + 2$.

• (a) Find $f(g(x))$ and $g(f(x))$.
• (b) Solve $f(g(x)) = g(f(x))$.

Solution

(a)

$$\begin{aligned} f\big(g(x)\big) &= (x + 2)^2 - 3(x + 2) \\ &= x^2 + 4x + 4 - 3x - 6 \\ &= x^2 + x - 2 \end{aligned}$$$$g\big(f(x)\big) = (x^2 - 3x) + 2 = x^2 - 3x + 2$$

(b)

$$x^2 + x - 2 = x^2 - 3x + 2 \quad\Rightarrow\quad 4x = 4 \quad\Rightarrow\quad x = 1$$

Check: $f(g(1)) = f(3) = 9 - 9 = 0$ and $g(f(1)) = g(-2) = 0$. ✓ So the two orders agree only at $x = 1$.

5. (Core) Let $f(x) = \dfrac{1}{x - 1}$ and $g(x) = x^2$. Find $f(g(x))$ and $g(f(x))$, and state the domain of each.

Solution

$$f\big(g(x)\big) = \frac{1}{x^2 - 1}$$

$f$ can’t accept $1$, so we need $x^2 \ne 1$: the domain is $\{x \in \mathbb{R} \mid x \ne \pm 1\}$.

$$g\big(f(x)\big) = \left(\frac{1}{x - 1}\right)^2 = \frac{1}{(x - 1)^2}$$

$f(x)$ needs $x \ne 1$, and $g$ accepts anything. The domain is $\{x \in \mathbb{R} \mid x \ne 1\}$.

6. (Core) Find functions $f$ and $g$ so that $h(x) = f(g(x))$. (There’s more than one answer; give the most natural one.)

• (a) $h(x) = (5x - 2)^3$
• (b) $h(x) = \sqrt{x^2 + 9}$
• (c) $h(x) = 2^{3x + 1}$
• (d) $h(x) = \dfrac{1}{(x + 4)^2}$

Solution

Look for the “inside” expression and call it $g(x)$.

(a) $g(x) = 5x - 2$ and $f(x) = x^3$.

(b) $g(x) = x^2 + 9$ and $f(x) = \sqrt{x}$.

(c) $g(x) = 3x + 1$ and $f(x) = 2^x$.

(d) $g(x) = x + 4$ and $f(x) = \dfrac{1}{x^2}$. (Another answer: $g(x) = (x + 4)^2$ and $f(x) = \dfrac{1}{x}$.)

7. (Core) Oil leaking from a tanker spreads in a circle on the water. The radius after $t$ minutes is $r(t) = 4t$ metres, and the area of a circle is $A(r) = \pi r^2$.

• (a) Find $A(r(t))$ and explain what it represents.
• (b) Find the area after $5$ minutes.
• (c) When does the area reach $2000$ m²?

Solution

(a) $A(r(t)) = \pi(4t)^2 = 16\pi t^2$. It’s the area of the slick, in square metres, $t$ minutes after the leak starts.

(b) $A(r(5)) = 16\pi(25) = 400\pi \approx 1256.64$ m².

(c) Solve $16\pi t^2 = 2000$ with $t \gt 0$:

$$t^2 = \frac{2000}{16\pi} \approx 39.79 \quad\Rightarrow\quad t \approx 6.31$$

The area reaches $2000$ m² after about $6.31$ minutes.

8. (Challenge) Let $f(x) = 2x + 3$. Find $g(x)$ if $f(g(x)) = 6x^2 - 1$.

Solution

$f$ doubles its input and adds $3$, so $f(g(x)) = 2g(x) + 3$. Set this equal to the given expression:

$$\begin{aligned} 2g(x) + 3 &= 6x^2 - 1 \\ 2g(x) &= 6x^2 - 4 \\ g(x) &= 3x^2 - 2 \end{aligned}$$

Check: $f(3x^2 - 2) = 2(3x^2 - 2) + 3 = 6x^2 - 1$. ✓

9. (Challenge) Let $f(x) = \dfrac{1}{x - 2}$. Find $f(f(x))$ in simplest form and state its domain.

Solution

$$f\big(f(x)\big) = \frac{1}{\dfrac{1}{x - 2} - 2}$$

Multiply the top and bottom by $(x - 2)$:

$$f\big(f(x)\big) = \frac{x - 2}{1 - 2(x - 2)} = \frac{x - 2}{5 - 2x}$$

Domain: the inside $f(x)$ needs $x \ne 2$. The outside $f$ can’t accept $2$, so we also need $\dfrac{1}{x - 2} \ne 2$, which means $x - 2 \ne \tfrac{1}{2}$, or $x \ne \tfrac{5}{2}$. (That matches the denominator $5 - 2x \ne 0$.)

$$\left\{x \in \mathbb{R} \;\middle|\; x \ne 2,\ x \ne \tfrac{5}{2}\right\}$$

Notice that $x = 2$ is excluded even though the simplified formula gives $\tfrac{0}{1} = 0$ there.