Chapter 14.1, Problem 2CE

(a)

To determine

To find: The image of $-3,6\text{and -6}$ if function $f:x=\left|x\right|$ .

Answer to Problem 2CE

The image of $-3,6\text{and -6}$ are $3,6$ and 6, respectively.

Explanation of Solution

Calculation:

The given function is $f\left(x\right)=\left|x\right|$ .

To find the image for the given values of $x$ , substitute the values of $x$ .

Consider $x=-3$ .

  $\begin{array}{c}f\left(-3\right)=\left|-3\right|\\ =3\end{array}$

Similarly,

Consider $x=6$ .

  $\begin{array}{c}f\left(6\right)=\left|6\right|\\ =6\end{array}$

and,

Consider $x=-6$ .

  $\begin{array}{c}f\left(-6\right)=\left|-6\right|\\ =6\end{array}$

Therefore, the image of $-3,6\text{and -6}$ are $3,6$ and 6, respectively.

(b)

To determine

To check: Whether the function $f\left(x\right)=\left|x\right|$ at $-3,6$ and $-6$ is one-to-one function.

Answer to Problem 2CE

No, the function $f\left(x\right)=\left|x\right|$ at $-3,6$ and $-6$ is not one-to-one function.

Explanation of Solution

Calculation:

Consider the given function $f\left(x\right)=\left|x\right|$ .

In part (a), the image of $-3,6$ and $-6$ are $3,6$ and 6, respectively. Since 6 and $-6$ both has image as 6. So, this is not one-to one function.

Therefore, the function $f\left(x\right)=\left|x\right|$ at $-3,6$ and $-6$ is not one-to-one function.