## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# 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.

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