Chapter 4.7, Problem 24PPS

To determine

To Graph :

Graph the function and state the domain and range.

Answer to Problem 24PPS

Domain is all real numbers.

Range = $\{y:y\ge 1\}$

Explanation of Solution

Given Information:

Given equation is :

  $y=g(x)=|2x-3|+1$

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 : $f(x)=\left|x\right|=\{\begin{array}{l}x;x>0\\ -x;x\le 0\end{array}$

  $\left|x\right|\ge 0$

Domain : All real numbers

Range : All non-negative real numbers

Graph :

  

Calculation :

  $\begin{array}{l}|2x-3|\ge 0\\ \Rightarrow g(x)=|2x-3|+1\ge 1\end{array}$

Since g(x) cannot be less than 1 , so the minimum point of the graph will be where | 2x -3|=0.

  $\begin{array}{l}|2x-3|=0\\ \Rightarrow \pm (2x-3)=0\\ \Rightarrow 2x-3=0\\ \Rightarrow 2x-3+3=0+\mathrm{3......................}[adding"3"each\_side]\\ \Rightarrow 2x=3\\ \Rightarrow \frac{2x}{2}=\frac{3}{2}\\ \Rightarrow x=\frac{3}{2}\end{array}$

Make a table for different value in the domain :

  $y=g(x)=|2x-3|+1=\{\begin{array}{l}2x-3+1;2x-3>0\\ -(2x-3)+1;2x-3\le 0\end{array}=\{\begin{array}{l}2x-2;x>\frac{3}{2}\\ 4-2x;x\le \frac{3}{2}\end{array}$

x	$2x-3$	$y=g(x)=|2x-3|+1$
0	-3	4
1	-1	2
1.5	0	1
2	1	2
3	3	4

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 =1 , the range is = $\left\{y:y\ge 1\right\}$