Chapter 4.7, Problem 21PPS

To determine

To Graph :

Graph the function and state the domain and range.

Answer to Problem 21PPS

Domain is all real numbers.

Range is non-negative real numbers = $\{y:y\ge 0\}$

Explanation of Solution

Given Information:

Given equation is :

  $y=f(x)=|\frac{1}{2}x-2|$

Concept Used :

Absolute Value Function:

An absolute value function is a function that contains an algebraic expression within absolute value symbols. The absolute value of a number is its distance from 0 on the number line. Parent Function : $f(x)=\left|x\right|=\{\begin{array}{l}x;x>0\\ -x;x\le 0\end{array}$

Domain : All real numbers

Range : All non-negative real numbers

Graph :

  

Calculation :

Since f(x) cannot be negative , so the minimum point of the graph will be where f(x)= 0.

  $\begin{array}{l}f(x)=|\frac{1}{2}x-2|=0\\ \Rightarrow f(x)=\pm (\frac{1}{2}x-2)=0\\ \Rightarrow \frac{1}{2}x-2=0\\ \Rightarrow \frac{1}{2}x-2+2=0+\mathrm{2......................}[adding"2"each\_side]\\ \Rightarrow \frac{1}{2}x=2\\ \Rightarrow \frac{1}{2}x(2)=2(2)\mathrm{.................................}[multiply"2"each\_side]\\ \Rightarrow x=4\end{array}$

Make a table for different value in the domain :

  $y=f(x)=|\frac{1}{2}x-2|=\{\begin{array}{l}\frac{1}{2}x-2;\frac{1}{2}x-2>0\\ -(\frac{1}{2}x-2);\frac{1}{2}x-2\le 0\end{array}=\{\begin{array}{l}\frac{1}{2}x-2;x>4\\ -(\frac{1}{2}x-2);x\le 4\end{array}$

x	$\frac{1}{2}x-2$	$y=f(x)=|\frac{1}{2}x-2|$
2	-1	1
3	-0.5	0.5
4	0	0
5	0.5	0.5
6	1	1

Plotting the values and joining, we get the required graph :

  

The x -coordinate values specify the domain of the function . Since the graph covers all possible values of x , the domain is all real numbers.

The y -coordiante values specify the range of the function . Since , the graph does not take any value less than y =0 or any negative value , the range is = $\left\{y:y\ge 0\right\}$