# `f(x) = x - 2tan^-1(x), [0, 4]` Find the absolute maximum and minimum values of f on the given interval

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Expert Answers

gsarora17 | Certified Educator

`f(x)=x-2tan^-1(x)`

differentiating,

`f'(x)=1-2/(x^2+1)`

`f'(x)=(x^2+1-2)/(x^2+1)`

`f'(x)=(x^2-1)/(x^2+1)`

Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers.

Now to find the critical numbers, solve for x for f'(x)=0.

`(x^2-1)/(x^2+1)=0`

`x^2-1=0 , x=+-1`

Now the critical number -1 is not in the interval (0,4).

So evaluate the function at the critical number 1 and at the end points of the interval (0,4).

`f(0)=0-2tan^-1(0)=0`

`f(4)=4-2tan^(-1)(4)=1.34836`

`f(1)=1-2tan^(-1)(1)=1-pi/2=-0.5708`

So, the function has **absolute minimum=1-`pi` /2 at x=1**