# 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).