Calculus (3rd Edition)

The maximum is $f(\pi/4) \approx-0.30349$ and the minimum is $f( 0 ) =-1$
Given $$y=\sqrt{2} \theta-\sec \theta, \quad\left[0, \frac{\pi}{3}\right]$$ Since \begin{align*} \frac{dy}{d\theta}&=\sqrt{2}-\sec \theta\tan \theta \end{align*} To find critical points \begin{align*} f'(t)&=0\\ \sqrt{2}-\sec \theta\tan \theta&=0 \end{align*} Then $\theta= \pi/4$. Because $\pi/4\in [0,\pi/3]$, then $f(x)$ has a critical point $\theta=\pi/4$ , since \begin{align*} f(0)&=-1 \\ f(\pi/3 )& \approx-0.519039\\ f(\pi/4)&\approx-0.30349 \end{align*} Then the maximum is $f(\pi/4) \approx-0.30349$ and the minimum is $f( 0 ) =-1$