Homework Help

with the function f(x)= `(2x)/(1+5x^2)^2`  ,  -2`<= x<=` the absolute max...

user profile pic

good2beme | Student, Undergraduate | Honors

Posted February 20, 2013 at 1:43 AM via web

dislike 0 like

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

the absolute max value? when x equals? the absolute min value? when x equals? 

1 Answer | Add Yours

user profile pic

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

Posted February 20, 2013 at 3:30 AM (Answer #1)

dislike 0 like

Given `f(x)=(2x)/((1+5x^2)^2)` for `-2<=x<=2` find the absolute max and min:

Since the function is continuous and on a closed interval, we are guaranteed both a maximum and a minimum. We must check any critical values as well as the endpoints:

The critical values are where the first derivative is zero or fails to exist.

We use the quotient rule to find the first derivative:

`f'(x)=(2(1+5x^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)`

The first derivative is continuous everywhere so we need only check its zeros:

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

---------------------------------------------------------------

The function's absolute maximum occurs at `x=sqrt(1/15)` while its absolute minimum occurs at `x=-sqrt(1/15)` regardless of the interval.

----------------------------------------------------------------

The graph:

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes