Homework Help: Find the maximum of 'arbitrary power' function

1. Oct 21, 2014

FlorenceC

1. If a and b are positive numbers, find the maximum value of f(x)=x^a(2-x)^b D={0<=x<=2}

attempt of solution
I did this question more intuitively.
So I first differentiated and found it to be axa-1(2-x)b + xa (2-x)bb-1
and I figured it will only be <0 when x>=2 and b is odd, so it will be >0 on the left which means maxima is at x=2 .

Is this right? and how do i prove it rigorously?

2. Oct 22, 2014

haruspex

Your differentiation is not quite right (sign error).

3. Oct 22, 2014

RUber

To find the zeros of the derivative, (once you have the right derivative), you will be able to factor out $x^{a-1}$ and $(2-x)^{b-1}$ terms from the derivative. These have zeros when a, b are not 1.
The remaining terms should give you a zero in terms of a and b.

4. Oct 22, 2014

RUber

Also, I think your logic is wrong for x=2 being a maximum. Plug 2 and 0 into the original function. Then over the remainder of the domain, you can see that f(x) > 0 for x in (0,2).