Chapter 4.6, Problem 12P

Find the range of each function for the given domain.

$f(x)=x^2+2;\left\{0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1\right\}$

To determine

To find: the range of given function for the given domain.

Answer to Problem 12P

The range of the given domain is

  $\left\{2,\frac{33}{16},\frac{9}{4},\frac{41}{16},3\right\}$

Explanation of Solution

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

Domain: $\left\{0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1\right\}$

the range by substitute the given domain in function,

For x=0,

  $\begin{array}{l}f\left(0\right)={\left(0\right)}^2+2\\ =2\end{array}$

For $x=\frac{1}{4}$ ,

  $\begin{array}{l}f\left(\frac{1}{4}\right)={\left(\frac{1}{4}\right)}^2+2\\ =\frac{1}{16}+2\\ =\frac{33}{16}\end{array}$

For $x=\frac{1}{2}$ ,

  $\begin{array}{l}f\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^2+2\\ =\frac{1}{4}+2\\ =\frac{9}{4}\end{array}$

For $x=\frac{3}{4}$

  $\begin{array}{l}f\left(\frac{3}{4}\right)={\left(\frac{3}{4}\right)}^2+2\\ =\frac{9}{16}+2\\ =\frac{41}{16}\end{array}$

For x=1

  $\begin{array}{l}f\left(1\right)=1^2+2\\ =1+2\\ =3\end{array}$

Therefore the range of the given domain is

  $\left\{2,\frac{33}{16},\frac{9}{4},\frac{41}{16},3\right\}$