University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.1 - Extreme Values of Functions - Exercises - Page 215: 25

Answer

$F(x)$ has an absolute maximum value of $-0.25$ at $x=2$ and an absolute minimum value of $-4$ at $x=0.5$ on the defined domain.
1543905619

Work Step by Step

$$F(x)=-\frac{1}{x^2}\hspace{1cm}0.5\le x\le2$$ 1) Find the derivative $F'(x)$: $$F'(x)=-\frac{(1)'\times x^2-1\times(x^2)'}{x^4}=-\frac{0\times x^2-2x}{x^4}$$ $$F'(x)=-\frac{(-2x)}{x^4}=\frac{2}{x^3}$$ Here there is no value of $x$ for which $F'(x)=0$. However, we can see that $F'(x)$ is undefined at $x=0$, which means $x=0$ is a critical point of $F(x)$. However, since the defined domain here is $[0.5,2]$, we would not take $x=0$ into consideration. 2) Evaluate $F(x)$ at the endpoints: - For $x=0.5$: $$F(0.5)=-\frac{1}{0.5^2}=-\frac{1}{0.25}=-4$$ - For $x=2$: $$F(2)=-\frac{1}{2^2}=-\frac{1}{4}=-0.25$$ This means the function $F(x)$ has an absolute maximum value of $-0.25$ at $x=2$ and an absolute minimum value of $-4$ at $x=0.5$ on the defined domain.
Small 1543905619
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.