Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 3 - Applications of Differentiation - 3.1 Exercises: 39

Answer

a) Minimum: (1, -1), Maximum: (-1, 3) b) Maximum: (3, 3) c) Minimum: (1, -1) d) Minimum: (1, -1)

Work Step by Step

Given: $f(x)=x^{2}-2x$ Find critical number(s). $f'(x)=2x-2$ (take the derivative of $f(x)$) $f'(x)=2(x-1) = 0$ (x is a critical number when $f'(x)=0$) $0 = x-1$ (divide both sides by 2, then add 1 to both sides) $x=1$ (there is a critical number when $x=1$) a) Given: $[-1, 2]$ Since the interval is closed, plug in both -1, 2 AND the critical number 1 to the function $f(x)$ $f(-1)=(-1)^{2}-2(-1)=3$ $f(1)=1^{2}-2(1)=-1$ $f(2)=2^{2}-2(2)=0$ The y value was smallest at $(1, -1)$ and biggest at $(-1, 3)$, so absolute minimum at $(1, -1)$ and absolute maximum at $(-1, 3)$ b) Given: $(1,3]$ The function is open at $x=1$ so we don't plug in $x=1$ to the function, even though it's a critical number. We still have to plug in the x value when the function is closed, when $x=3$ $f(3)=3^{2}-2(3)=3$ So absolute maximum at $(3, 3)$ c) Given: $(0,2)$ The interval is open interval, just plug in the critical number 1 to the function. $f(x)=1^{2}-2(1)=-1$ So absolute minimum at $(1, -1)$ d) [1,4) The interval is open at $x=4$ so we don't plug in $x=4$ to the function, we just need to plug in the critical number/closed interval 1 to the function. $f(x)=1^{2}-2(1)=-1$ So absolute minimum at $(1, -1)$
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