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: 52

Answer

$f(3)=125$ is the absolute maximum value. $f(0)=-64$ is the absolute minimum value

Work Step by Step

The Closed Interval Method 1) Function $f$ must be continuous on the closed interval $[a, b].$ 2) Find the values of $f$ at the critical points in $(a,b)$ and the values of $f$ at the endpoints. 3) Compare the values. - The largest is the absolute maximum value. - The smallest is the absolute minimum value. $$f(t)=(t^2-4)^3,\hspace{1.5cm}[-2, 3]$$ First, $f(t)$ is continuous on interval $[-2,3]$, so the Closed Interval Method can be applied here. 1) Find critical points of $f$ $$f'(t)=3(t^2-4)^2(t^2-4)'$$ $$f'(t)=3(t^2-4)^22t$$ $$f'(t)=6t(t^2-4)^2.$$ Therefore, $$f'(t)=0$$ $$6t(t^2-4)^2=0$$ $$t=0\hspace{0.5cm}or\hspace{0.5cm}t^2=4$$ $$t=0\hspace{0.5cm}or\hspace{0.5cm}t=\pm2$$ These values all belong to the domain of $f$, so they are all critical points. However, only $t=0$ and $t=2$ lie in the interval $(-2, 3)$, so only these two will move to the next sections. 2) Calculate the values of $f$ at critical points and endpoints. - Critical points: $t=0$ and $t=2$ $$f(0)=(0^2-4)^3=(-4)^3=-64$$ $$f(2)=(2^2-4)^3=0^3=0$$ - End points: $t=-2$ and $t=3$ $$f(-2)=[(-2)^2-4]^3=0^3=0$$ $$f(3)=(3^2-4)^3=5^3=125$$ 3) Comparison Among the results: - $f(3)=125$ is the largest, so $f(3)=125$ is the absolute maximum value. - $f(0)=-64$ is the smallest, so $f(0)=-64$ is the absolute minimum value.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.