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: 55

Answer

$f(4)\approx2.413$ is the absolute maximum value. $f(\frac{\sqrt 3}{9})\approx-0.385$ 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. $$f(t)=t-\sqrt[3]t,\hspace{1.5cm}[-1, 4]$$ $$f(t)=t-t^{1/3}$$ First, $f(t)$ is continuous on the interval $[-1,4]$, so the Closed Interval Method can be applied here. 1) Find critical points of $f$ $$f'(t)=1-\frac{1}{3}t^{-2/3}$$ $$f'(t)=1-\frac{1}{3\sqrt[3]{t^2}}$$ We can easily see that $f'(t)$ does not exist when $t=0$. $t=0$ lies in the domain of $f$, so it is a critical point. Also, $$f'(t)=0$$ $$\frac{1}{3\sqrt[3]{t^2}}=1$$ $$\sqrt[3]{t^2}=\frac{1}{3}$$ $$t^2=\frac{1}{27}$$ $$t=\pm\frac{1}{3\sqrt 3}=\pm\frac{\sqrt 3}{9}$$ These values also are critical points. After all, we've found 3 critical points. They all lie in interval $(-1, 4)$ given, so they will all move to the next sections. 2) Calculate the values of $f$ at critical points and endpoints. - Critical points: $t=0$ and $t=\pm\frac{\sqrt 3}{9}$ $$f(0)=0-\sqrt[3]0=0$$ $$f(\frac{\sqrt 3}{9})=\frac{\sqrt 3}{9}-\sqrt[3]{\frac{\sqrt 3}{9}}\approx-0.385$$ $$f(\frac{-\sqrt 3}{9})=\frac{-\sqrt 3}{9}-\sqrt[3]{\frac{-\sqrt 3}{9}}\approx0.385$$ - End points: $t=-1$ and $t=4$ $$f(-1)=-1-\sqrt[3]{-1}=-1-(-1)=0$$ $$f(4)=4-\sqrt[3]4\approx2.413$$ 3) Comparison Among the results: - $f(4)\approx2.413$ is the largest, so $f(4)\approx2.413$ is the absolute maximum value. - $f(\frac{\sqrt 3}{9})\approx-0.385$ is the smallest, so $f(\frac{\sqrt 3}{9})\approx-0.385$ is the absolute minimum value.
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