Chapter 1.9, Problem 47E

(a)

To determine

To find: inverse of the function of f .

Answer to Problem 47E

The inverse of the given function is $f^{-1}(x)=\sqrt{4-x^2}$ .

Explanation of Solution

Given information:

The function $f(x)=\sqrt{4-x^2},0\le x\le 2$

Calculation:

Let the function $y=f(x)=\sqrt{4-x^2}$

  $y=\sqrt{4-x^2}$

  $\begin{array}{l}{y}^2=4-x^2\\ x^2=4-y^2\\ x=\sqrt{4-y^2}\\ f^{-1}(y)=\sqrt{4-y^2}\\ f^{-1}(x)=\sqrt{4-x^2}\end{array}$

Hence, the inverse of the given function is $f^{-1}(x)=\sqrt{4-x^2}$ .

(b)

To determine

To sketch: the graphs both the functions f and $f^{-1}.$

Explanation of Solution

Given information:

The function $f(x)=\sqrt{4-x^2},0\le x\le 2$

Graph:

The graph of the functions $f(x)=\sqrt{4-x^2},0\le x\le 2$ and $f^{-1}(x)=\sqrt{4-x^2}$ is drawn below.

(c)

To determine

To describe: the relation between the graphs of f and $f^{-1}.$

Explanation of Solution

Given information:

The function $f(x)=\sqrt{4-x^2},0\le x\le 2$

Consider the graph of part $(b)$

The graph of the $f^{-1}$ is the reflection of graph f about the line $y=x.$

In this case both the functions are equal.

(d)

To determine

To state: The domains and ranges of the functions f and $f^{-1}$ .

Explanation of Solution

Given information:

The function $f(x)=\sqrt{4-x^2},0\le x\le 2$

Domain is the set of input values for which the function is defined.

Range is the set of defined output values.

From the graphs of part b,

Consider $f(x)=f^{-1}(x)=\sqrt{4-x^2}$ is

Domain: $0\le x\le 2$ , Range: $0\le y\le 2$

Hence, the domains of f and $f^{-1}$ are $[0,2]$ .