Answer
The absolute maximum is $f(4) = 1.35$
The absolute minimum is $f(1) = 1-\frac{\pi}{2}$
Work Step by Step
$f(x) = x-2~tan^{-1}~x$
We can find the points when $f'(x) = 0$:
$f(x) = x-2~tan^{-1}~x$
$f'(x) = 1-\frac{2}{x^2+1} = 0$
$\frac{x^2+1}{x^2+1}-\frac{2}{x^2+1} = 0$
$\frac{x^2-1}{x^2+1} = 0$
$x^2-1 = 0$
$x = -1, 1$
We can find the value of $f(1)$ and the values of the function at the endpoints of the interval $[0, 4]$:
$f(1) =(1)-2~tan^{-1}~(1) = 1-\frac{\pi}{2}$
$f(0) =(0)-2~tan^{-1}~(0) = 0$
$f(4) =(4)-2~tan^{-1}~(4) = 1.35$
The absolute maximum is $f(4) = 1.35$
The absolute minimum is $f(1) = 1-\frac{\pi}{2}$
