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f(x)= 2x/(1+5x^2)^2, -2`<=x<=1` The absolute max value is? and occurs at x...

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good2beme | Student, Undergraduate | (Level 1) Honors

Posted February 21, 2013 at 2:49 AM via web

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f(x)= 2x/(1+5x^2)^2, -2`<=x<=1`

The absolute max value is? and occurs at x equals? The absolute min value is? and this occurs at x equals?

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted February 21, 2013 at 5:52 AM (Answer #1)

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Since the function is continuous on [-2,1] there is a maximum and a minimum. The extrema can occur at critical points or the endpoints of the interval. Critical points occur where the first derivative is zero or fails to exist.

First we compute the first derivative:

Using the quotient rule we get `f'(x)=d/(dx)[(2x)/((1+5x^2)^2)]`

`=((1+5x^2)^2(2)-(2x)(2)(1+5x^2)(10x))/((1+5x^2)^4)`

`=((1+5x^2)(2(1+5x^2)-40x^2))/((1+5x^2)^4)`

`=(2-30x^2)/((1+5x^2)^3)`

`f'(x)=0==>2-30x^2=0==>x=+-sqrt(1/15)`

We evaluate f at `-2,-sqrt(1/15),sqrt(1/15),` and 1.

`f(-2)~~-.0091`

`f(-sqrt(1/15))~~-.2905`

`f(sqrt(1/15))~~.2905`

`f(1)=.0bar(5)`

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The maximum on the interval occurs at `x=sqrt(1/15)` with a value of `f(sqrt(1/15))~~.2905` . ** `-sqrt(1/15)~~-.2582`

The minimum on the interval occurs at `x=-sqrt(1/15)` with a value of `f(-sqrt(1/15))~~-.2905` ** `sqrt(1/15)~~.2582`

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The graph:

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