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: 41

Answer

$$\theta= n\pi \hspace{4mm} \forall n\in Z $$ are the critical numbers of $f$.

Work Step by Step

How to find the critical numbers of a function $f$ according to the definition 1) Find all numbers of $c$ satisfying either $f'(c)=0$ or $f'(c)$ does not exist. 2) See that all the $c$ we get lie in the domain of $f$ $(D_f)$ or not. - If $c$ lies in $D_f$, $c$ is a critical number of $f$. - If $c$ does not lie in $D_f$, $c$ is not a critical number of $f$. $$f(\theta)=2\cos\theta+\sin^2\theta$$ $D_f=R$ 1) Now, we find $f'(\theta)$ $$f'(\theta)=-2\sin\theta+2\sin\theta(\sin\theta)'$$ $$f'(\theta)=-2\sin\theta+2\sin\theta\cos\theta$$ $$f'(\theta)=-2\sin\theta(1-\cos\theta)$$ We find $f'(\theta)=0$, $$f'(\theta)=0$$ $$-2\sin\theta(1-\cos\theta)=0$$ $$\sin\theta=0\hspace{0.5cm}or\hspace{0.5cm}\cos\theta=1$$ $$\theta=\pm n\pi\hspace{0.5cm}or\hspace{0.5cm}\theta=\pm2n\pi$$ which means, $$\theta= n\pi \hspace{4mm} \forall n\in Z $$ These values belong to $D_f$, so they are critical number of $f$. There are no values of $\theta$ that can make $f'(\theta)$ not exist. We conclude that $$\theta= n\pi \hspace{4mm} \forall n\in Z $$ are the critical numbers of $f$.
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