
Concept explainers
To sketch: The function g(x)=|x2−1|−|x2−4|.

Explanation of Solution
Given:
The function g(x)=|x2−1|−|x2−4|.
Result used:
Even function:
A function is even if and only if g(x)=g(−x) for all x in the domain of g.
Definition used:
The absolute value of x is defined as, |x|={x if x≥0−x if x<0.
Calculation:
Consider the function g(x)=|x2−1|−|x2−4|
Now, |x2−1|={x2−1 if x≥11−x2 if x<1
|x2−4|={x2−4 if x≥24−x2 if x<2
The function can be split into different cases as follows:
Case (i):
For 0≤x≤1
g1(x)=x2+1−x2−4g1(x)=−3
To sketch: Use the online graphing calculator and draw the graph of the function: g1(x)=−3 as shown below in Figure 1.
Case (ii):
For 1≤x<2
g2(x)=x2−1+x2−4g2(x)=2x2−5
To sketch: Use the online graphing calculator and draw the graph of the function: g2(x)=2x2+5 as shown below in Figure 2
Case (iii):
For x≥2
g3(x)=x2−1−x2+4g3(x)=3
To sketch: Use the online graphing calculator and draw the graph of the function: g3(x)=3 as shown below in Figure 3.
Case (iv):
For −1<x≤0
g1(−x)=x2+1−x2−4g1(−x)=−3
To sketch: Use the online graphing calculator and draw the graph of the function: g1(−x)=−3 as shown below in Figure 4.
Case (v):
For −2<x≤−1
g2(−x)=x2−1+x2−4g2(−x)=2x2−5
To sketch: Use the online graphing calculator and draw the graph of the function: g2(−x)=2x2+5 as shown below in Figure 5.
Case (vi):
For x≤−2
g3(−x)=x2−1+x2−4g3(−x)=3
To sketch: Use the online graphing calculator and draw the graph of the function: g3(−x)=3 as shown below in Figure 6.
Combine all the graphs in figure 1, 2,3,4,5 and 6 as shown below in Figure 7.
To sketch: Use the online graphing calculator and draw the graph of the function: g(x)=|x2−1|−|x2−4| as shown below in Figure 7.
Chapter 1 Solutions
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