Find the maximum of 'arbitrary power' function

[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 ax^a-1(2-x)^b + x^a (2-x)b^b-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?

 

[haruspex]

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 ax^a-1(2-x)^b + x^a (2-x)b^b-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?

Your differentiation is not quite right (sign error).

 

[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.

 

[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).