## Calculus (3rd Edition)

The minimum is $f(\pi/4) =1-\dfrac{\pi}{2}$ and the maximum is $f( 0) =0$
Given $$y=\tan x-2 x, \quad[0,1]$$ Take the derivative \begin{align*} \frac{dy}{dx}&=\sec^2x-2 \end{align*} Find the critical points \begin{align*} f'(\theta)&=0\\ \sec^2x-2 &=0 \end{align*} Then $x=\pi/4 \in [0,1]$ and \begin{align*} f(0)&=0 \\ f(\pi/4)&=1-\frac{\pi}{2}\\ f(1)&=\tan (1)-2 \end{align*} Then the minimum is $f(\pi/4) =1-\dfrac{\pi}{2}$ and the maximum is $f( 0) =0$