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: 51


Absolute Max:$(-2,33)$ Absolute Min:$(2,-31)$

Work Step by Step

Original equation:$ f(x)=3x^{4}-4x^{3}-12x^{2}+1$ on the interval $[-2,3]$ 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: $f′(x)=12x^{3}-12x^{2}-24x$ Then, we set this derivative to zero and solve for x using elementary algebra. We see that the critical point occurs when x=2 $12x^{3}-12x^{2}-24x=0$ $x^{3}-x^{2}-2x=0$ $x(x^{2}-x-2)=0$ $x(x+1)(x-2)=0$ $x=0, -1, 2$ Now we have five possible points of where a absolute minimum or maximum may occur: the critical points $x=0, -1, 2$, as well as the two endpoints of the interval, where $x=-2,3$. 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. $f(-2)=3(-2)^{4}-4(-2)^{3}-12(-2)^{2}+1=33$ $f(-1)=3(-1)^{4}-4(-1)^{3}-12(-1)^{2}+1=-4$ $f(0)=3(0)^{4}-4(0)^{3}-12(0)^{2}+1=1$ $f(2)=3(2)^{4}-4(2)^{3}-12(2)^{2}+1=-31$ $f(3)=3(3)^{4}-4(3)^{3}-12(3)^{2}+1=28$ We see that the minimum y value occurs when x=2 and the maximum y value occurs when x=-2. Therefore, we have an absolute maximum at $(-2,33)$ and an absolute minimum at $(2,-31)$.
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