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: 47


Absolute maximum at $(2, 16)$ Absolute minimum at $(5, 7)$

Work Step by Step

Original equation: $f(x)=12+4x-x^{2} $ on the interval $[0,5]$ 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: $f'(x)= 4-2x$ Then, we set this derivative to zero and solve for x using elementary algebra. We see that the critical point occurs when $x=2$ $4-2x = 0$ $-2x=-4$ $x=2$ Now we have three possible points of where a absolute minimum or maximum may occur: the critical point $x=2$, as well as the two endpoints of the interval, where $x=0,5$. 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. $f(0) = 12+4(0)-(0)^{2} = 12$ $f(2) = 12+4(2)-(2)^{2} = 16$ $f(5) = 12+4(5)-(5)^{2} = 7$ Of the three values, we see that the minimum y value occurs when $x=5$ and the maximum y value occurs when $x=2$. Therefore, we have an absolute maximum at $(2, 16)$ and an absolute minimum at $(5, 7)$.
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