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: 48


Absolute maximum at (3,113) Absolute minimum at (0,5)

Work Step by Step

Original equation:$ f(x)=5+54x−2x^{3}$ on the interval $[0,4]$ 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: $f′(x)=54-6x^{2}$ Then, we set this derivative to zero and solve for x using elementary algebra. We see that the critical point occurs when x=2 $54-6x^{2}=0$ $6x^{2}=54$ $x^{2}=9$ $x=3, -3$ In our given interval $[0,4]$, the only critical point we will consider is when $x=3$ as $x=-3$ is outside of our given interval. Now we have three possible points of where a absolute minimum or maximum may occur: the critical point $x=3$, as well as the two endpoints of the interval, where $x=0,4$. 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. $f(0)=5+54(0)−2(0)^{3}=5$ $f(3)=5+54(3)−2(3)^{3}=113$ $f(4)=5+54(4)−2(4)^{3}=93$ Of the three values, we see that the minimum y value occurs when x=0 and the maximum y value occurs when x=3. Therefore, we have an absolute maximum at $(3,113)$ and an absolute minimum at $(0,5)$.
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