Chapter 3.8, Problem 11CYU

To determine

To graph the function $g(x)=\left|\frac{1}{2}x\right|$ and determine its domain and range and its relation with the graph of the function $f(x)=\left|x\right|$ .

Answer to Problem 11CYU

Domain: $x\in ℝ$

Range: $y\in \left[0,\infty \right)$

Explanation of Solution

Given:

Function: $g(x)=\left|\frac{1}{2}x\right|$

Concept used:

The general form of an absolute value function is $f(x)=a\left|x-h\right|+k$ .

The graph of $f(x)=a\left|x-h\right|+k$ is a “V” with vertex $\left(h,k\right)$ , slope $m=a$ on the right side of the vertex $\left(x>h\right)$ and slope $m=-a$ on the left side of the vertex $\left(x<h\right)$ .

The graph of $f(x)=-a\left|x-h\right|+k$ is an upside-down “V” with vertex $\left(h,k\right)$ , slope $m=-a$ for $\left(x>h\right)$ and slope $m=a$ for $\left(x<h\right)$ .

Calculation for graph:

Consider $g(x)=\left|\frac{1}{2}x\right|=\frac{1}{2}\left|x\right|,$

Compare with the general form:

  $g(x)=a\left|x-h\right|+k$

Here, the value of $a=\frac{1}{2}$ , so the graph opens upwards with a slope of $\frac{1}{2}$ .

The value of $h=0$ and the value of $k=0$ , so the vertex of the graph is $\left(0,0\right)$ .

Values of x	Values of g (x)
0	$0$
1	$\frac{1}{2}$
-1	$\frac{1}{2}$
2	$1$
-2	$1$

By taking different values of x , the graph can be plotted.

Graph:

The graph of $g(x)=\left|\frac{1}{2}x\right|$ :

  

Interpretation:

From the graph, it can be observed that the horizontal extent of the graph is from $-\infty$ to $\infty$ , so the domain of $g(x)$ is $\left(-\infty ,\infty \right)$ .

The vertical extent of the graph is from $0$ to $\infty$ , so the range of $g(x)$ is $\left[0,\infty \right)$ .

Comparison:

Now, comparing the graph of $g(x)$ with the graph of $f(x)=\left|x\right|$ .

Graph of $f(x)=\left|x\right|:$

  

The vertex, domain and range of both the graphs are $(0,0)$ , $\left(-\infty ,\infty \right)$ and $\left[0,\infty \right)$ respectively.

Therefore, both the graphs have same vertex, same domain and same range.

Conclusion:

Therefore, the domain of $g(x)=\left|\frac{1}{2}x\right|$ is $x\in ℝ$ and range is $y\in \left[0,\infty \right)$ .