Chapter 4.5, Problem 2E

(a)

To determine

To find: $f\left(\left[1,3\right]\right).$

Answer to Problem 2E

  $f\left(\left[1,3\right]\right)=\left[2,10\right]$

Explanation of Solution

Given information: $f\left(x\right)=x^2+1$

Definition used:

Let $f:A\to B$ and let $X\subseteq A$ and $Y\subseteq B$ the image of X or image set of X is $f(x)=\left\{y\in B:y=f(x)\text{ for some }x\in X\right\}$

Calculation:

The graph of the given function f is shown below:

  

From the graph, it is clear that on the subset $\left[1,3\right]$ the function $f\left(x\right)$ is increasing.

The minimum value of function f will be: $f\left(1\right)=1+1=2$

The maximum value of function f will be: $f\left(3\right)=3^2+1=9+1=10$

Since $f\left(x\right)$ is a continuous function it reaches every value in that interval.

∴ $f\left(\left[1,3\right]\right)=\left[2,10\right]$

(b)

To determine

To find: $f\left(\left[-1,0\right]\cup \left[2,4\right]\right)$

Answer to Problem 2E

  $f\left(\left[-1,0\right]\right)\cup f\left(\left[2,4\right]\right)$

  $=\left[1,2\right]\cup \left[5,17\right]$

Explanation of Solution

Given information: $f\left(x\right)=x^2+1$

Theorem used:

Let $f:A\to B,C\text{ and }D$ be subsets of A.

Then, $f\left(C\cup D\right)=f\left(C\right)\cup f\left(D\right)$

Calculation:

From the graph, it is clear that on the subset $\left[-1,0\right]$ , the function $f\left(x\right)$ is decreasing

The minimum value of function f will be: $f\left(0\right)=0+1=1$

The maximum value of function f will be: $f\left(-1\right)={\left(-1\right)}^2+1=2$

On the subset $\left[2,4\right]$ , the function $f\left(x\right)$ is increasing

The minimum value of function f will be: $f\left(2\right)=2^2+1=5$

The maximum value of function f will be: $f\left(4\right)=4^2+1=17$

  $\begin{array}{c}f\left(\left[-1,0\right]\right)\cup f\left(\left[2,4\right]\right)=f\left(\left[-1,0\right]\right)\cup f\left(\left[2,4\right]\right)\\ =\left[1,2\right]\cup \left[5,17\right]\end{array}$

(c)

To determine

To find: $f^{-1}\left(\left[-1,1\right]\right)$

Answer to Problem 2E

  $f^{-1}\left(\left[-1,1\right]\right)=\left\{0\right\}$

Explanation of Solution

Given information: $f\left(x\right)=x^2+1$

Definition used:Let $f:A\to B$ and let $X\subseteq A$ and $Y\subseteq B$ , the inverse image of Y is $f^{-1}\left(Y\right)=\left\{x\in A:f\left(x\right)=y\right\}$

Calculation:

Since $f\left(x\right)=x^2+1$

  $\ge 1$ , the only possible value in the interval $\left[-1,1\right]$ is $\left\{1\right\}$ .

That is $f^{-1}\left(\left\{1\right\}\right)=0$ [since $f\left(0\right)=1$ ]

(d)

To determine

To find: $f^{-1}\left(\left[-2,3\right]\right)$

Answer to Problem 2E

  $f^{-1}\left(\left[-2,3\right]\right)=\left[-\sqrt{2},\sqrt{2}\right]$

Explanation of Solution

Given information: $f\left(x\right)=x^2+1$

Definition used: Let $f:A\to B$ and let $X\subseteq A$ and $Y\subseteq B$ , the inverse image of Y is $f^{-1}\left(Y\right)=\left\{x\in A:f\left(x\right)=y\right\}$

Calculation:

Since $f\left(x\right)=x^2+1$

  $\ge 1$ , the only possible values in the interval $\left[-2,3\right]$ is $\left[1,3\right]$

That is, for

  $\begin{array}{l}1\le f\left(x\right)\le 3\\ \Rightarrow 1\le x^2+1\le 3\\ \Rightarrow 0\le x^2\le 2\end{array}$

∴The possible values of x are $-\sqrt{2}\le x\prec 0\text{ , 0}\le x\le \sqrt{2}$

∴ $f^{-1}\left(\left[-2,3\right]\right)=\left[-\sqrt{2},\sqrt{2}\right]$

(e)

To determine

To find: $f^{-1}\left(\left[5,10\right]\right)$

Answer to Problem 2E

  $f^{-1}\left(\left[5,10\right]\right)=\left[-3,-2\right]\cup \left[2,3\right]$

Explanation of Solution

Given information: $f\left(x\right)=x^2+1$

Formulas used:Let $f:A\to B$ and let $X\subseteq A$ and $Y\subseteq B$ , the inverse image of Y is $f^{-1}\left(Y\right)=\left\{x\in A:f\left(x\right)=y\right\}$

Calculation:

For, $5\le f\left(x\right)\le 10$

  $\begin{array}{l}\Rightarrow 5\le x^2+1\le 10\\ \Rightarrow 4\le x^2\le 9\end{array}$

∴The possible values of x are $-3\le x\le -2\text{ , }2\le x\le 3$

∴ $f^{-1}\left(\left[5,10\right]\right)=\left[-3,-2\right]\cup \left[2,3\right]$

(f)

To determine

To find: $f^{-1}\left(\left[-1,5\right]\cup \left[17,26\right]\right)$

Answer to Problem 2E

  $f^{-1}\left(\left[-1,5\right]\cup \left[17,26\right]\right)=\left[-5,-4\right]\cup \left[-2,2\right]\cup \left[4,5\right]$

Explanation of Solution

Given information: $f\left(x\right)=x^2+1$

Formulas used:

Let $f:A\to B$ and let $X\subseteq A$ and $Y\subseteq B$ , the inverse image of Y is $f^{-1}\left(Y\right)=\left\{x\in A:f\left(x\right)=y\right\}$

  $f^{-1}\left(C\cup D\right)=f^{-1}\left(C\right)\cup f^{-1}\left(D\right)$

Calculation:

To calculate $f^{-1}\left(\left[-1,5\right]\right)$ :

Since $f\left(x\right)=x^2+1$

  $\ge 1$ , the only possible values in the interval $\left[-1,5\right]$ is $\left[1,5\right]$

That is, for $1\le f\left(x\right)\le 5$

  $\begin{array}{l}1\le x^2+1\le 5\\ \Rightarrow 0\le x^2\le 4\\ \Rightarrow -2\le x\le 2\end{array}$

∴ $f^{-1}\left(\left[-1,5\right]\right)=\left[-2,2\right]$

To calculate $f^{-1}\left(\left[17,26\right]\right)$ :

  $\begin{array}{l}17\le x^2+1\le 26\\ \Rightarrow 16\le x^2\le 25\end{array}$

∴The possible values of x are $-5\le x\le -4\text{ , 4}\le x\le 5$

  $f^{-1}\left(\left[17,26\right]\right)=\left[-5,-4\right]\cup \left[4,5\right]$

  $\begin{array}{c}{f}^{-1}\left(\left[-1,5\right]\cup \left[17,26\right]\right)=f^{-1}\left(\left[-1,5\right]\right)\cup f^{-1}\left(\left[17,26\right]\right)\\ =\left[-5,-4\right]\cup \left[-2,2\right]\cup \left[4,5\right]\end{array}$