Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.1 - Maximum and Minimum Values - 4.1 Exercises - Page 284: 50


Absolute Maximum: $(0,5)$ Absolute Minimum: $(-3,-76)$

Work Step by Step

Original equation:$ f(x)=x^{3} - 6x^{2}+5$ on the interval $[-3,5]$ To find the absolute maximum and minimum, we first need to find all the critical points of the function, as the local extrema might be the absolute extrema. Critical points of a function occur when a function's derivative is undefined or the derivative is zero. In this case, we see that the function is a polynomial, which means it is defined on all intervals from $(−\infty,\infty)$. Therefore, we need to set the derivative to zero. Finding the derivative of the function through the power rule: $f′(x)=3x^{2}-12x$ Then, we set this derivative to zero and solve for x using elementary algebra. We see that the critical point occurs when x=0,4 $3x^{2}-12x=0$ $x^{2}-4x=0$ $x(x-4)=0$ $x=0, 4$ Now we have four possible points of where a absolute minimum or maximum may occur: the critical points $x=0,4$, as well as the two endpoints of the interval, where $x=-3,5$. To find which one of these is an absolute extrema vs a local extrema, we plug these x values back into the original equation and see what we get. $f(-3)=(-3)^{3} - 6(-3)^{2}+5=-76$ $f(0)=(0)^{3} - 6(0)^{2}+5=5$ $f(4)=(4)^{3} - 6(4)^{2}+5=-27$ $f(5)=(5)^{3} - 6(5)^{2}+5=-20$ Of the three values, we see that the minimum y value occurs when x=-3 and the maximum y value occurs when x=0. Final Answer: Absolute Maximum: $(0,5)$ Absolute Minimum: $(-3,-76)$
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