## University Calculus: Early Transcendentals (3rd Edition)

$g(x)$ has the absolute maximum value of $2$ at $x=-\frac{\pi}{3}$ and the absolute minimum value of $1$ at $x=0$ in the defined domain.
$$g(x)=\sec x\hspace{1cm}-\frac{\pi}{3}\le x\le\frac{\pi}{6}$$ 1) Find the derivative $g'(x)$: $$g'(x)=\sec x\tan x=\frac{\tan x}{\cos x}$$ - $g'(x)=0$ when $\tan x x=0$, which is when $x=k\pi$ for $k\in Z$. - $g'(x)$ is undefined when $\cos x=0$, which is when $x=\frac{\pi}{2}+k\pi$ $(k\in Z)$. All these values of $x$ are the critical points of $g(x)$. However, among all these values, only $x=0$ lies in the defined domain $[-\pi/3,\pi/6]$. So only the critical point $x=0$ would be taken into consideration here. 2) Evaluate $g'(x)$ at the critical points and endpoints: - For $x=-\pi/3$: $$g\Big(-\frac{\pi}{3}\Big)=\sec\Big(-\frac{\pi}{3}\Big)=\frac{1}{\cos\Big(-\frac{\pi}{3}\Big)}=\frac{1}{\frac{1}{2}}=2$$ - For $x=0$: $$g(0)=\sec0=\frac{1}{\cos0}=\frac{1}{1}=1$$ - For $x=\pi/6$: $$g\Big(\frac{\pi}{6}\Big)=\sec\Big(\frac{\pi}{6}\Big)=\frac{1}{\cos\frac{\pi}{6}}=\frac{1}{\frac{\sqrt3}{2}}=\frac{2}{\sqrt3}=\frac{2\sqrt3}{3}$$ This means $g(x)$ has the absolute maximum value of $2$ at $x=-\frac{\pi}{3}$ and the absolute minimum value of $1$ at $x=0$ in the defined domain.