Inverse with domain restriction (HL)

Recall that inverse functions only exist for invertible (one-to-one) functions. Otherwise, the functions needs to be split into domains of invertible functions, called branches. The desired domain may be specified in the question or by the context.

Also if question says an inverse exists, then it exists without having to break up into branches. This allows you to find domain and range of the inverse without having to find the inverse itself.

For inverse of quadratics, a simple $\pm$ is sufficient to describe the two branches.

Example: Given $f(x) = 9x^2 + 12x - 3, x \leq -1$. Find $f^{-1}(x)$ and state the domain.

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Completing the square:

We have $a = 9, b = 12, c = -3$ with minimum at ${x = -\frac{12}{2\cdot9}= f\left(-\frac23\right)}$. It is a minimum as $a > 0$.

$$\begin{align*} f\left(-\frac23\right) &= 9\left(-\frac23\right)^2 + 12\left(-\frac23\right) - 3\\ &= 4 - 8 - 3 \\ &= -7 \end{align*}$$

Hence, $f(x) = 9\left(x + \frac23\right)^2 - 7$. We begin the process of finding the inverse

$$\begin{align*} x &= 9\left(y + \frac23\right)^2 - 7\\ x + 7 &= 9\left(y + \frac23\right)^2 \\ \frac{x+7}9 &= \left(y + \frac23\right)^2 \\ \pm \sqrt{\frac{x+7}9} &= y + \frac23 \\ \frac{-2 \pm \sqrt{x+7}}3 &= y \end{align*}$$

We are given an endpoint at $x = -1$.

$$\begin{align*} f(-1) &= 9(-1)^2 + 12(-1) - 3 \\ &= 9 - 12 - 3 \\ &= -6 \end{align*}$$

Since $f(-1) < -\frac23$, we keep only the minus branch. Since $f(x) \geq -6$, and range of $f$ becomes domain of $f^{-1}$, the final answer is

$$f^{-1}(x) = \frac{-2 - \sqrt{x+7}}3,\, x \geq -6 \qed$$

Inverse of trig functions (bonus)

Probably not required, but it’s technically within the realm of possibilities.

Prerequisite: transformations of functions

$\arcsin$ and $\arctan$ are increasing functions, from an increasing interval of $\sin$ and $\tan$, while $\arccos$ is a decreasing function, from a decreasing interval of $\cos$.

If $\sin$, $\cos$, or $\tan$ is vertically reflected across $x$-axis, intervals of increasing function becomes intervals of decreasing functions, vice versa. $\cos$ is unaffected by a horizontal reflection across $y$-axis, but the other two are, with similar increasing to decreasing changes.

So if looking for the inverse function of a decreasing portion of $\sin$ or increasing portion of $\cos$, then we need to multiply the inverse by $-1$.

Conceptually, we identify transformations, then convert vertical stretches and translations to horizontal stretches and translations in the inverse, and vice versa.

Example: Given

$$f(x) = -3\cos\left(2\left(x + \frac{\pi}{3}\right)\right) + 1, x \in \left[\frac{7\pi}{6}, \frac{5\pi}{3}\right]$$

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

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See method at transformation of trig functions, there is a maximum at ${\left(\frac{7\pi}{6}, 4\right)}$ and a minimum at $\left(\frac{5\pi}{3}, -2\right)$. This confirms $f$ is decreasing.

A simplistic approach for the inverse function yields

$$x \mapsto \frac{\arccos \left(\frac{x-1}{-3}\right)}{2} - \frac{\pi}{3}$$

While $f$ is decreasing, it also had a vertical reflection, meaning it was transformed from an increasing branch in $x\mapsto \cos x$. As $\arccos$ is for a decreasing branch of $\cos$, we need to multiply the inverse by $-1$. We can ignore the vertical shift ${-\frac{\pi}{3}}$ for now.

$$g(x) = \frac{\arccos \left(\frac{x-1}{-3}\right)}{-2}$$

Time to figure out that vertical shift in the final answer.

$$g(4) = \frac{\arccos \left(\frac{4-1}{-3}\right)}{-2} = -\frac{\pi}{2}$$

$$\frac{7\pi}{6} - \left(-\frac{\pi}{2}\right) = \frac{5\pi}{3}$$

This means we need to add $\frac{5\pi}{3}$ to ${-\frac{\pi}{2}}$ to get $\frac{7\pi}{6}$.

$$f^{-1}(x) = \frac{\arccos \left(\frac{4-1}{-3}\right)}{-2} + \frac{5\pi}{3}, x \in \left[-2, 4\right] \qed$$