Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.3 - How Derivatives Affect the Shape of a Graph - 4.3 Exercises - Page 301: 20

Answer

Using the First Derivative Test or the Second Derivative Test, we can determine that $f$ has a local maximum at $x = 0$ and $f$ has a local minimum at $x = 2$

Work Step by Step

$f(x) = \frac{x^2}{x-1}$ First Derivative Test: $f'(x) = \frac{(2x)(x-1)-(x^2)(1)}{(x-1)^2}$ $f'(x) = \frac{x^2-2x}{(x-1)^2} = 0$ $x^2-2x = 0$ $x(x-2) = 0$ $x = 0, 2$ As $x \to 0^-,$ then $f'(x) \gt 0$ As $x \to 0^+,$ then $f'(x) \lt 0$ $f$ has a local maximum at $x = 0$ As $x \to 2^-,$ then $f'(x) \lt 0$ As $x \to 2^+,$ then $f'(x) \gt 0$ $f$ has a local minimum at $x = 2$ Second Derivative Test: $f'(0) = 0$ $f'(2) = 0$ $f''(x) = \frac{(2x-2)(x-1)^2-(x^2-2x)(2)(x-1)}{(x-1)^4}$ $f''(x) = \frac{(2x-2)(x-1)-(x^2-2x)(2)}{(x-1)^3}$ $f''(x) = \frac{2x^2-4x+2-2x^2+4x}{(x-1)^3}$ $f''(x) = \frac{2}{(x-1)^3}$ At $x = 0,~~f''(x) \lt 0$ $f$ has a local maximum at $x = 0$ At $x = 2,~~f''(x) \gt 0$ $f$ has a local minimum at $x = 2$
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