Finding Critical points (derivatives)?

Question

Hi.

I'm trying to find the critical points for the function :

f(x)= x sqrt(x - (x^2))

I don't seem to know what to do after using the product rule with chain rule...

I got:

[(x(1-2x))/(2 sqrt(x-x^2))] + sqrt[x-x^2]

Can anyone show me the steps on how to proceed with this? I would be super grateful.


Thanks!!

PS: I know the answer is 0 and 0.75 --> I need to know the steps.

yes, I know you have to set the derivative to zero, but the simplifying part has me stumped.

The part where you add the fractions specifically, see http://answers.yahoo.com/question/index;_ylt=ApBmzjxKPlF2MwtARRHSQoYCxgt.?qid=20070725214326AAA3RbP for my specific problem.

Please try to simplify for me and show work.

Thanks

I see what I did wrong now.

Thank you garry i I will try to remember to choose you as the best answer

Answer 1

f(x) = x (x - x²)^(1/2)
f `(x) =
(x - x²)^(1/2) + (1/2) (x - x²)^(-1/2) (1 - 2x) (x)
(1/2) (x - x²)^(-1/2) [ 2 (x - x²) + x (1 - x) ]
(1/2) (x - x²)^(- 1/2) [ 3x - 4x² ]
x (3 - 4x) / 2 ( x - x² )^(1/2)
For turning points f `(x) = 0:-
x = 0 , x = 3 / 4 (required answer)

Answer 2

well since you have already found the derivative by using the product rule and the chain rule, then all you do is set your derivative equal to zero and solve for x. Simplify your derivative as much as you can before you begin to solve for x. That will help you out alot.