   Chapter 1, Problem 5P ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Sketch the graph of the function f(x) = |x2 – 4| x | + 3|.

To determine

To sketch: The function f(x)=|x24|x|+3|.

Explanation

Given:

The function is f(x)=|x24|x|+3|.

Result used:

Even function:

A function is even if and only if f(x)=f(x) for all x in the domain of f.

Calculation:

Consider the function f(x)=|x24|x|+3|

Observe that x2=(x)2and |x|=|x|

Therefore, f(x)=f(x) and thus the given function is an even function.

Thus, for all x0.

f(x)=|x24x+3|=(x1)(x3)

The function can be split into four different cases as follows:

Case (i):

For 0x<1,and x3

f1(x)=x24x+3

To sketch: Use the online graphing calculator and draw the graph of the function: f1(x)=x24x+3 as shown below in Figure (1)

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 