Help me solve the algebra in this calculus question..?

Question

Calculus question: Solving for absolute minimum and absolute maximum?
For the value of f(x) = x/(x^2 +1) on the given interval [0,2]

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

did I find the derivative correctly??? and now what do I do to set it to zero. I am totally confused !!! HELP!! Thanks




Could you show me how to solve for the +/- 1 by eliminating the denominator. Once I have that I can solve for the min. and max. ...it's the algebra that driving me crazy!!! Thanks!!

I guess I'm not clear on what I don't understand how do I get from :
this
(x^2+1-2x^2)/(x^2+1)^2 to
to
x^2 = 1 ???

that part in between is the part I'm confused about; once I can do that then I believe I know how to find the min and max.... Thanks !!!

Answer 1

Yes you did work out the derivative correctly
f'(x) = (1 - x^2) / (x^2 + 1)^2

f' = 0 when x^2 = 1
or x = +/- 1
Note that for f'(x) to be zero, you need the numerator to be zero, so forget about the denominator for now.

Only x = 1 is in our interval

f(1) = 1 / (1^2 + 1) = 1/2
f(0) = 0
f(2) = 2 / 5

So it's clear that f(1) = 1/2 is a maximum turning point.
From that maximum the function decreases, so the minimum value will be at one of the ends, in this case f(0) = 0 is the minimum.

**EDIT**
You have f'(x) = (x^2 + 1 - 2x^2) / (x^2 + 1)^2
= (1 - x^2) / (x^2 + 1)^2
Now to find the turning points (max or min), we need to set
f'(x) = 0
Thus (1 - x^2) / (x^2 + 1)^2 = 0
The only way for that to happen is for the numerator to be zero, ie
1 - x^2 = 0
x^2 = 1
x = +/- 1 will be the location of the turning points.

Answer 2

Hello

Since you set it = 0 then you can multiply both sides by the denominator and it then drops out. You have left x^2 +1 -2x^2 or x^2 +1 = 0 now subtract 1 from both sides giving us x^2 = 1 take the square root of both sides we have x = +/- 1
Since it is from 0 to 2 you use 1 and put it into the original equation along with say 0.9 and 1.1 to see if 1 is a max or min.
Hope This Helps!!

Answer 3

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

we need to find something that makes f'(x) zero inorder to find the critical points.
based on observation, the denominator can never be zero.

therefore, -x^2 + 1 = 0
x^2 = 1
x = +1, -1

next sub the value 0, 1 ,2 into fx)
-1 is not taken as it is not in the interval

f(0) = 0
f(1) = 0.5
f(2) = 0.4

hence the absolute maximum is at (1, 0.5)
absolute minimum is at (0, 0) in the interval of [0, 2]