   Chapter 4.3, Problem 27E ### 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 a function that satisfies all of the given conditions.f′(0) = f′(2) = f′(4) = 0,f′(x) > 0 if x < 0 or 2 < x < 4,f′(x) < 0 if 0 < x < 2 or x > 4,f″(x) > 0 if 1 < x < 3, f″(x) < 0 if x < 1 or x > 3

To determine

To sketch: The graph of the function which satisfy the following conditions that f(0)=f(2)=f(4)=0 , f(x)>0 if x<0 or 2<x<4 , f(x)<0 if 0<x<2 or x>4 , f(x)>0 if 1<x<3 , f(x)<0 if x<1 or x>3 .

Explanation

Definition used:

Increasing/Decreasing Test:

“(a) If f(x)>0 on an interval, then f is increasing on that interval.

(b) If f(x)<0 on an interval, then f is decreasing on that interval”.

Concavity Test:

“(a) If f(x)>0 for all x in I, then the graph of f is concave upward on I.

(b) If f(x)<0 for all x in I, then the graph of f is concave downward on I”.

Graph:

The conditions f(0)=f(2)=f(4)=0 means that horizontal tangents exists at x=0,2,4 .

From the above definition the condition f(x)>0 if x<0 or 2<x<4 indicates that f is increasing on the intervals

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