# Show that the derivative of f(x) = (x-a)m (x-b)n vanishes at some point between a & b

Question: Show that the derivative of f(x) = (x-a)m (x-b)n vanishes at some point between a and b if m and n are positive integers.
My attempt:
f(x) = (x-a)m (x-b)n
f '(x) = m(x-a)m-1 (x-b)n + n(x-a)m (x-b)n-1
f '(x) = [(x-a)n-1 (x-b)n-1 ] [(m)(x-b) +(n)(x-a)]
And this is as far as I got.

easily can be shown by rolle's theorem

We cant use that yet :(

You have the expression for the derivative. Set it equal to zero and you will find one value of x for which the derivative is always zero, namely x=(mb+na)/(m+n). You then have to show that it is always between a and b if m and n are positive integers. Actually, m and n can be any reals greater than or equal to 1.

Why can't you use Rolle's theorem if you understand it?

HallsofIvy
Homework Helper

Since this question has nothing to do with "Differential Equations", I am moving it to Calculus and Beyond Homework.

HallsofIvy
Homework Helper

Question: Show that the derivative of f(x) = (x-a)m (x-b)n vanishes at some point between a and b if m and n are positive integers.
My attempt:
f(x) = (x-a)m (x-b)n
f '(x) = m(x-a)m-1 (x-b)n + n(x-a)m (x-b)n-1
f '(x) = [(x-a)n-1 (x-b)n-1 ] [(m)(x-b) +(n)(x-a)]
And this is as far as I got.
Set that equal to 0 and solve for x:
[(x-a)n-1 (x-b)n-1 ] [(m)(x-b) +(n)(x-a)]= 0
Since we are looking at x between a and b, x- a and x- b are not 0 so we can divide by them. That leaves m(x- b)+ n(x- a)= mx- mb+ nx- na= (m-n)x- mb-na= 0. Solve that for x and show that it lies between a and b.

Thanks guys! I ended up using Rolles theorem. I'm still waiting to see if he accepts it as a valid approach.