Answer
The minimum is $ f(\pi/4) =1-\dfrac{\pi}{2}$ and the maximum is $f( 0) =0$
Work Step by Step
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 $f( 0) =0$
