Student Question

# `f(x) = (2x)/(x^2 + 1), [-2,2]` Find the absolute extrema of the function on the closed interval.

## Expert Answers

Given: `f(x)=(2x)/(x^2+1)[-2,2]`

Find the critical values for x by setting the derivative equal to zero and solving for the x value(s).

`f'(x)=[(x^2+1)(2)-(2x)(2x)]/(x^2+1)^2=0`

`2x^2+2-4x^2=0`

`-2x^2+2=0`

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

`x=1,x=-1`

Plug in the critical x value(s) and the endpoints of the closed interval into the f(x) function.

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

`f(-2)=-.8`

`f(-1)=-1`

`f(1)=1`

`f(2)=-.8`

Examine the f(x) values to determine the absolute extrema.

The absolute maximum is the point (1, 1).

The absolute minimum is the point (-1, -1).

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You need to evaluate the absolute extrema of the function, hence, you need to differentiate the function with respect to x, using the quotient rule, such that:

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

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

`f'(x) = (2x^2 + 2 - 4x^2)/((x^2+1)^2)`

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

You need to solve for x the equation f'(x) = 0

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

Factoring out 2 yields:

`2(1 - x^2) = 0` `=> 1 - x^2 = 0 => (1 - x)(1 + x) = 0 => x = 1` and `x = -1`

Both solutions belongs to the interval [-2.2].

Hence,evaluating the absolute extrema of the function, yields that the function reaches it's extrema at x = 1 and x  = -1.

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