Chapter 7.1, Problem 53AYU

In Problems 63-70, find the inverse function $f^{-1}$ of each function $f$. Find the range of f and the domain and range of $f^{-1}$.

$f\left(x\right)=5\sin x+2;-\frac{\pi }{2}\le x\le \frac{\pi }{2}$

To determine

To find: The inverse function $f^{-1}$ of function $f\left(x\right)=5\sin x+2$; $-\pi /2\le x\le \pi /2$.

Range of $f$.

Domain and range of $f^{-1}$.

Answer to Problem 53AYU

Solution:

$f^{-1}\left(x\right)={\sin }^{-1}\left(x-2\right)/5$

The range of $f$ is $\left\{y/-3\le y\le 3\right\}$ or $\left[-3,3\right]$ using interval notation.

The domain of $f^{-1}$ becomes the range of $f=\left[-3,3\right]$.

The range of $f^{-1}$ becomes the domain of $f=\left[-\pi /2,\pi /2\right]$.

Explanation of Solution

Given:

$f\left(x\right)=5\sin x+2;-\pi /2\le x\le \pi /2$

Calculation:

The function $f\left(x\right)=5\sin x+2$ is a one to one function, so has inverse function.

To calculate the inverse function,

Replace $f\left(x\right)$ by variable $y$.

$y=5\text{sin}x+2$

Interchange $x$ and $y$.

$x=5\sin y+2$

Subtract 2 on both sides,

$x–2=5\sin y$

Solve for $y$.

$\sin y=\left(x–2\right)/5$   -----1

$y={\sin }^{-1}\left(x-2\right)/5$

The inverse function.

$f^{-1}\left(x\right)={\sin }^{-1}\left(x-2\right)/5$

To find the range of $f$, solve $y$ in equation 1.

$y=5\sin x+2$ for $\sin x$ and use the fact $-1\le \sin x\le 1$.

$\sin x=\left(y-2\right)/5$

$-1\le \left(y-2\right)/5\le 1$

$-5\le \left(y-2\right)\le 1$

$-3\le y\le 3$

The range of $f$ is $\left\{y/-3\le y\le 3\right\}$ or $\left[-3,3\right]$ using interval notation.

The domain of $f^{-1}$ is the range of $f=\left[-3,3\right]$.

The range of $f^{-1}$ is the domain of $f=\left[-\pi /2,\pi /2\right]$.