Composition, Inverses, and Transformations

Composition gives you a new way to look at two ideas you already know. An inverse is the function that, composed with $f$, takes every number back to where it started. And every transformation you’ve done, like stretches and shifts, is really a composition with a simple linear function. Trig functions here use radians, and $\log x$ means $\log_{10} x$.

Key ideas

A function and its inverse cancel

If $f$ sends $a$ to $b$, its inverse sends $b$ back to $a$. Put the two machines in a row and the number comes out unchanged:

$$a \;\xrightarrow{\;f\;}\; b \;\xrightarrow{\;f^{-1}\;}\; a$$

So composing a function with its inverse, in either order, maps every number onto itself:

$$f^{-1}\big(f(x)\big) = x \;\text{ for every } x \text{ in the domain of } f \qquad f\big(f^{-1}(x)\big) = x \;\text{ for every } x \text{ in the domain of } f^{-1}$$

For example, if $f(2) = 9$, then $f^{-1}(f(2)) = f^{-1}(9) = 2$.

Checking inverses by composition

Two functions $f$ and $g$ are inverses of each other exactly when

$$f\big(g(x)\big) = x \quad\text{and}\quad g\big(f(x)\big) = x$$

for every $x$ in the right domains. You need to check both orders.

Watch the domains

The identities only hold on the right domains, and this matters when a domain has been restricted.

• $f(x) = x^2$ with $x \ge 0$ has inverse $f^{-1}(x) = \sqrt{x}$. Then $\sqrt{x^2} = x$ only for $x \ge 0$. For every real $x$, $\sqrt{x^2} = \lvert x \rvert$: for example, $\sqrt{(-3)^2} = 3$, not $-3$.
• $f(x) = 10^x$ and $f^{-1}(x) = \log x$. Then $\log(10^x) = x$ for every real $x$, but $10^{\log x} = x$ only for $x \gt 0$, since $\log x$ needs a positive input.

Transformations are compositions

Let $g(x) = A(x + B)$, a linear function. Composing it with $f$ in the two different orders gives the two kinds of transformations.

$g$ on the inside: horizontal changes.

$$f\big(g(x)\big) = f\big(A(x + B)\big)$$

This is $y = f(k(x - d))$ with $k = A$ and $d = -B$: a horizontal stretch or compression by a factor of $\dfrac{1}{\lvert A \rvert}$ (and a reflection in the $y$-axis if $A \lt 0$), then a translation $B$ units left (right if $B \lt 0$).

$g$ on the outside: vertical changes.

$$g\big(f(x)\big) = A\big(f(x) + B\big) = A f(x) + AB$$

This is a vertical stretch by a factor of $\lvert A \rvert$ (and a reflection in the $x$-axis if $A \lt 0$), then a translation of $AB$ units up. Notice that the vertical shift is $AB$, not $B$: the $+B$ happens first, and then it gets stretched too.

So “inside” changes act on $x$ (horizontally), and “outside” changes act on $y$ (vertically), just as in combined transformations.

Comparing the two orders

With $g(x) = 2(x - 1)$, so $A = 2$ and $B = -1$:

$f(x)$	$f(g(x))$: horizontal	$g(f(x))$: vertical
$x^2$	$\big(2(x - 1)\big)^2 = 4(x - 1)^2$: vertex $(1, 0)$	$2(x^2 - 1) = 2x^2 - 2$: vertex $(0, -2)$
$\sin x$	$\sin\big(2(x - 1)\big)$: period $\pi$, shifted $1$ right	$2(\sin x - 1) = 2\sin x - 2$: amplitude $2$, axis $y = -2$
$\log x$	$\log\big(2(x - 1)\big)$: asymptote $x = 1$	$2(\log x - 1) = 2\log x - 2$: asymptote $x = 0$

For $f(x) = x^2$ and $g(x) = 2(x - 1)$: $f(g(x))$ moves the vertex sideways, while $g(f(x))$ moves it down.

Worked examples

Example 1: Numbers going in and coming back

Let $f(x) = 3x - 7$.

• (a) Find $f^{-1}(x)$.
• (b) Evaluate $f^{-1}(f(4))$ and $f(f^{-1}(-1))$ step by step.
• (c) Show that $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$ for all $x$.

Solution.

(a) Swap $x$ and $y$ in $y = 3x - 7$: $x = 3y - 7$, so $y = \dfrac{x + 7}{3}$. That is, $f^{-1}(x) = \dfrac{x + 7}{3}$.

(b) $f(4) = 12 - 7 = 5$, and $f^{-1}(5) = \dfrac{12}{3} = 4$. So $f^{-1}(f(4)) = 4$.

$f^{-1}(-1) = \dfrac{6}{3} = 2$, and $f(2) = 6 - 7 = -1$. So $f(f^{-1}(-1)) = -1$.

Each number came back to where it started.

(c)

$$f^{-1}\big(f(x)\big) = \frac{(3x - 7) + 7}{3} = \frac{3x}{3} = x$$ $$f\big(f^{-1}(x)\big) = 3\left(\frac{x + 7}{3}\right) - 7 = (x + 7) - 7 = x$$

Example 2: Checking inverses by composition

Show that $f(x) = \dfrac{x - 1}{x + 2}$ and $g(x) = \dfrac{2x + 1}{1 - x}$ are inverses of each other.

Solution. Compose in both orders. Each time, simplify the top and the bottom over a common denominator.

$$\begin{aligned} f\big(g(x)\big) &= \frac{\dfrac{2x + 1}{1 - x} - 1}{\dfrac{2x + 1}{1 - x} + 2} = \frac{\dfrac{2x + 1 - (1 - x)}{1 - x}}{\dfrac{2x + 1 + 2(1 - x)}{1 - x}} \\ &= \frac{3x}{3} = x \end{aligned}$$ $$\begin{aligned} g\big(f(x)\big) &= \frac{2\left(\dfrac{x - 1}{x + 2}\right) + 1}{1 - \dfrac{x - 1}{x + 2}} = \frac{\dfrac{2(x - 1) + (x + 2)}{x + 2}}{\dfrac{(x + 2) - (x - 1)}{x + 2}} \\ &= \frac{3x}{3} = x \end{aligned}$$

Both compositions give $x$ (for $x \ne -2$ in $g(f(x))$ and $x \ne 1$ in $f(g(x))$), so $f$ and $g$ are inverses.

Quick check: $f(4) = \dfrac{3}{6} = 0.5$ and $g(0.5) = \dfrac{2}{0.5} = 4$. ✓

Example 3: When the domain matters

Let $f(x) = x^2 - 4$ for $x \ge 0$, with inverse $f^{-1}(x) = \sqrt{x + 4}$.

• (a) Simplify $f^{-1}(f(x))$ and $f(f^{-1}(x))$, and state where each equals $x$.
• (b) What goes wrong if you put $x = -3$ into the formula for $f^{-1}(f(x))$?

Solution.

(a)

$$f^{-1}\big(f(x)\big) = \sqrt{(x^2 - 4) + 4} = \sqrt{x^2} = \lvert x \rvert$$

This equals $x$ for $x \ge 0$, which is exactly the domain of $f$. ✓

$$f\big(f^{-1}(x)\big) = \left(\sqrt{x + 4}\right)^2 - 4 = (x + 4) - 4 = x$$

This works for $x \ge -4$, the domain of $f^{-1}$ (and the range of $f$). ✓

(b) $(-3)^2 - 4 = 5$ and $\sqrt{5 + 4} = 3$, so you get $3$, not $-3$. That’s not a contradiction: $-3$ isn’t in the restricted domain of $f$, so the identity was never promised for it.

Example 4: Transformations as compositions

Let $g(x) = 3(x - 2)$. For $f(x) = x^2$, $f(x) = \sin x$ and $f(x) = \log x$, find $f(g(x))$ and $g(f(x))$ and describe each as a transformation of $f$.

Solution. Here $A = 3$ and $B = -2$.

$f(x) = x^2$.

• $f(g(x)) = \big(3(x - 2)\big)^2 = 9(x - 2)^2$: a horizontal compression by a factor of $\tfrac{1}{3}$, then a translation $2$ units right. Vertex $(2, 0)$.
• $g(f(x)) = 3(x^2 - 2) = 3x^2 - 6$: a vertical stretch by a factor of $3$, then a translation $AB = -6$, so $6$ units down. Vertex $(0, -6)$.

$f(x) = \sin x$.

• $f(g(x)) = \sin\big(3(x - 2)\big)$: period $\tfrac{2\pi}{3}$, translated $2$ units right. The range stays $\{y \in \mathbb{R} \mid -1 \le y \le 1\}$.
• $g(f(x)) = 3(\sin x - 2) = 3\sin x - 6$: amplitude $3$, axis $y = -6$, period still $2\pi$. The range is $\{y \in \mathbb{R} \mid -9 \le y \le -3\}$.

$f(x) = \log x$.

• $f(g(x)) = \log\big(3(x - 2)\big)$: compressed horizontally by $\tfrac{1}{3}$ and shifted $2$ right. The domain is $\{x \in \mathbb{R} \mid x \gt 2\}$ and the vertical asymptote is $x = 2$. By the laws of logarithms, this also equals $\log 3 + \log(x - 2)$: for logarithms, a horizontal compression is the same as a vertical shift (up $\log 3 \approx 0.48$).
• $g(f(x)) = 3(\log x - 2) = 3\log x - 6$: stretched vertically by $3$ and shifted $6$ down. The domain is still $\{x \in \mathbb{R} \mid x \gt 0\}$, with asymptote $x = 0$. It passes through $(1, -6)$ and $(10, -3)$.

In every case, $g$ on the inside changed things horizontally, and $g$ on the outside changed things vertically.

Common mistakes

Checking only one composition. $f(g(x)) = x$ alone isn’t enough to prove that $f$ and $g$ are inverses in general. Check both orders.

Ignoring the domain. $\sqrt{x^2}$ is $\lvert x \rvert$, not $x$. The identity $f^{-1}(f(x)) = x$ only holds for $x$ in the domain of $f$, so a restricted domain matters.

Assuming $10^{\log x} = x$ for every $x$. $10^{\log x} = x$ needs $x \gt 0$; for $x = -5$, $\log(-5)$ doesn’t exist. (But $\log(10^x) = x$ works for every real $x$.)

Getting the direction of the horizontal shift wrong. $f(A(x + B))$ shifts the graph $B$ units left, the opposite of the sign you see. $\sin\big(3(x - 2)\big)$ moves $2$ units right.

Forgetting that the vertical shift gets stretched. $g(f(x)) = A(f(x) + B)$ moves the graph by $AB$, not $B$. With $g(x) = 3(x - 2)$, $g(f(x)) = 3f(x) - 6$, a shift of $6$ down, not $2$ down.

Confusing $f^{-1}$ with $\dfrac{1}{f}$. $f^{-1}(f(x)) = x$ is about undoing. $\dfrac{1}{f(x)} \cdot f(x) = 1$ is a different fact about reciprocals.

Practice

1. (Warm-up) $f$ is a function with an inverse, and $f(5) = 12$. Find:

• (a) $f^{-1}(f(5))$
• (b) $f(f^{-1}(12))$
• (c) $f^{-1}(12)$

Solution

(a) $f(5) = 12$, then $f^{-1}(12) = 5$. So $f^{-1}(f(5)) = 5$.

(b) $f^{-1}(12) = 5$, then $f(5) = 12$. So $f(f^{-1}(12)) = 12$.

(c) $f^{-1}(12) = 5$.

2. (Warm-up) Show that $f(x) = 4x + 1$ and $g(x) = \dfrac{x - 1}{4}$ are inverses by composition.

Solution

$$f\big(g(x)\big) = 4\left(\frac{x - 1}{4}\right) + 1 = x - 1 + 1 = x$$$$g\big(f(x)\big) = \frac{(4x + 1) - 1}{4} = \frac{4x}{4} = x$$

Both give $x$, so they’re inverses.

3. (Warm-up) Simplify each expression.

• (a) $10^{\log 7}$
• (b) $\log\left(10^{3.2}\right)$
• (c) $\left(\sqrt{11}\right)^2$

Solution

$10^x$ and $\log x$ are inverses, and so are $x^2$ (for $x \ge 0$) and $\sqrt{x}$. Each pair cancels:

(a) $7$ (b) $3.2$ (c) $11$

4. (Core) Are $f(x) = 2x - 6$ and $g(x) = \dfrac{x}{2} - 6$ inverses? If not, find the correct inverse of $f$.

Solution

$$f\big(g(x)\big) = 2\left(\frac{x}{2} - 6\right) - 6 = x - 12 - 6 = x - 18$$

That’s not $x$, so they are not inverses. To find $f^{-1}$: $x = 2y - 6$ gives $y = \dfrac{x + 6}{2} = \dfrac{x}{2} + 3$.

Check: $f\left(\dfrac{x}{2} + 3\right) = x + 6 - 6 = x$ ✓ and $\dfrac{2x - 6}{2} + 3 = x - 3 + 3 = x$. ✓

5. (Core) Show that $f(x) = \dfrac{1}{x - 3}$ and $g(x) = \dfrac{1}{x} + 3$ are inverses, and state the domain of each.

Solution

$$f\big(g(x)\big) = \frac{1}{\left(\frac{1}{x} + 3\right) - 3} = \frac{1}{\;\frac{1}{x}\;} = x$$$$g\big(f(x)\big) = \frac{1}{\;\frac{1}{x - 3}\;} + 3 = (x - 3) + 3 = x$$

Both give $x$, so they’re inverses. Domain of $f$: $\{x \in \mathbb{R} \mid x \ne 3\}$. Domain of $g$: $\{x \in \mathbb{R} \mid x \ne 0\}$. (Each domain is the other’s range.)

6. (Core) Let $f(x) = x^2 + 2$ for $x \le 0$.

• (a) Find $f^{-1}(x)$ and its domain.
• (b) Verify that $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$ on the right domains.

Solution

(a) Swap: $x = y^2 + 2$, so $y = \pm\sqrt{x - 2}$. The range of $f^{-1}$ must be the domain of $f$, $y \le 0$, so take the negative root:

$$f^{-1}(x) = -\sqrt{x - 2}, \qquad \{x \in \mathbb{R} \mid x \ge 2\}$$

(b) For $x \le 0$:

$$f^{-1}\big(f(x)\big) = -\sqrt{(x^2 + 2) - 2} = -\sqrt{x^2} = -\lvert x \rvert = x$$

The last step works because $x \le 0$, so $\lvert x \rvert = -x$. For $x \ge 2$:

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

7. (Core) Let $f(x) = \sin x$ and $g(x) = 2(x + 1)$. Find $f(g(x))$ and $g(f(x))$, and describe each as a transformation of $y = \sin x$. State the range of each.

Solution

$f(g(x)) = \sin\big(2(x + 1)\big)$: a horizontal compression by a factor of $\tfrac{1}{2}$ (period $\pi$), then a translation $1$ unit left. The range is $\{y \in \mathbb{R} \mid -1 \le y \le 1\}$.

$g(f(x)) = 2(\sin x + 1) = 2\sin x + 2$: a vertical stretch by a factor of $2$, then a translation $2$ units up. The range is $\{y \in \mathbb{R} \mid 0 \le y \le 4\}$.

8. (Challenge) Let $f(x) = \log x$ and $g(x) = 10(x + 2)$.

• (a) Find $f(g(x))$, and use a law of logarithms to show it is a vertical translation of $y = \log(x + 2)$.
• (b) Find $g(f(x))$ and state its domain.

Solution

(a)

$$f\big(g(x)\big) = \log\big(10(x + 2)\big) = \log 10 + \log(x + 2) = 1 + \log(x + 2)$$

So $y = \log(10(x + 2))$, which is $y = \log x$ compressed horizontally by a factor of $\tfrac{1}{10}$ and then shifted $2$ left, is the same graph as $y = \log(x + 2)$ translated up $1$ unit. The domain is $\{x \in \mathbb{R} \mid x \gt -2\}$.

(b) $g(f(x)) = 10(\log x + 2) = 10\log x + 20$. The domain is $\{x \in \mathbb{R} \mid x \gt 0\}$.

9. (Challenge) Let $f(x) = x^2$.

• (a) Find all linear functions $g(x) = A(x + B)$ with $f(g(x)) = 9x^2 - 12x + 4$.
• (b) Show that $h(x) = \dfrac{x + 5}{x - 1}$ is its own inverse by finding $h(h(x))$.

Solution

(a) $9x^2 - 12x + 4 = (3x - 2)^2$, so we need $\big(g(x)\big)^2 = (3x - 2)^2$. That means $g(x) = 3x - 2$ or $g(x) = -(3x - 2) = -3x + 2$. In the form $A(x + B)$:

$$g(x) = 3\left(x - \tfrac{2}{3}\right) \qquad\text{or}\qquad g(x) = -3\left(x - \tfrac{2}{3}\right)$$

(The second one includes a reflection in the $y$-axis, which doesn’t change the even function $x^2$.)

(b) Multiply the top and bottom by $(x - 1)$:

$$h\big(h(x)\big) = \frac{\dfrac{x + 5}{x - 1} + 5}{\dfrac{x + 5}{x - 1} - 1} = \frac{(x + 5) + 5(x - 1)}{(x + 5) - (x - 1)} = \frac{6x}{6} = x$$

So $h(h(x)) = x$ for $x \ne 1$, which means $h$ undoes itself: $h^{-1} = h$.