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: 40

Answer

$$\theta=\frac{\pi}{3}\pm2n\pi\hspace{0.5cm}and\hspace{0.5cm}\theta=\frac{2\pi}{3}\pm2n\pi\hspace{4mm} \forall n\in Z$$ are critical numbers of $g$.

Work Step by Step

How to find the critical numbers of a function $f$ according to definition 1) Find all numbers of $c$ satisfying whether $f'(c)=0$ 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 $f$. $$g(\theta)=4\theta-\tan\theta$$ $\tan\theta$ would be defined if $\cos\theta\ne0$ (for $\tan\theta=\frac{\sin\theta}{\cos\theta}$), which means $\theta\ne\frac{\pi}{2}\pm n\pi$ for all $n \in Z$. Therefore, $$D_g=\{\theta | \theta\ne\frac{\pi}{2}\pm n\pi \hspace{2mm} \forall n\in Z\}.$$ 1) Now, we find $g'(\theta)$ $$g'(\theta)=4-\frac{1}{\cos^2\theta}$$ We find $g'(\theta)=0$, $$g'(\theta)=0$$ $$\frac{1}{\cos^2\theta}=4$$ $$\cos^2\theta=\frac{1}{4}$$ $$\cos\theta=\pm\frac{1}{2}$$ $$\theta=\frac{\pi}{3}\pm2n\pi\hspace{0.5cm}or\hspace{0.5cm}\theta=\frac{2\pi}{3}\pm2n\pi \hspace{4mm} \forall n\in Z$$ All of these values are in $D_g$, so they are critical numbers of $g$. Also, looking at $$g'(\theta)=4-\frac{1}{\cos^2\theta}$$ reveals that for $\cos\theta=0$, $g'(\theta)$ would not exist. However, all values of $\theta$ satisfying $\cos\theta=0$ do not belong to $D_g$. Therefore, they are not critical numbers of $g$. We conclude that only $$\theta=\frac{\pi}{3}\pm2n\pi\hspace{0.5cm}and\hspace{0.5cm}\theta=\frac{2\pi}{3}\pm2n\pi \hspace{4mm} \forall n\in Z$$ are critical numbers of $g$.
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