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

### Textbook Question

Chapter 4, 4.1 - Problem 62 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`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

function has no absolute maximum , see graph

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