Absolute Extrema Homework

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...

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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