Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises - Page 180: 41


$f'(x)=4x^{3}-6x^{2}+2x$ The graphs of f'(x) correctly shows the increasing and decreasing nature of f(x), as well as all the vertices.

Work Step by Step

$f(x)=x^{4}-2x^{3}+x^{2}$ using the power rule, ($\frac{d}{dx}$[$x^{n}$]=n$x^{n-1}$) we can find that $f'(x)=4x^{3}-6x^{2}+2x$ When you compare the graphs of f(x) and f'(x), they make sense. This is because from the left (negative), moving right, f(x) is decreasing until x=0. f'(x) is negative here and at x=0, crosses the x-axis. f(x) then increases from 0 to 1/2, and then decreases from 1/2 to 1. Through this, f'(x) is positive, zero, then negative, just as it should be. At x=1, f(x) has a vertex, meaning f(x) should have a zero, which it does. After x=1, f(x) increases and f'(x) is positive.
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