# Calculus derivative of given function vanishes at some point between a and b

1. Oct 10, 2012

### tachyon_man

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.

2. Oct 10, 2012

### chiro

Hey kylem1994 and welcome to the forums.

Try re-writing m(x-b) + n(x-a) as a parametrization where a corresponds to t = 0 and b corresponds to t = 1 and show that there is a solution of t in between 0 and 1 where this whole term equals 0 after a re-parameterization.

3. Oct 10, 2012

### tachyon_man

So you're saying use the mean value theorem? I'm confused on how that would show that there's a point where the derivative 'vanishes'. What I'm thinking its its a discontinuous graph at a certain point but just proving it is tough.

4. Oct 10, 2012

### chiro

If m and n are positive, you won't need to worry about discontinuities since f'(x) is continuous (and you can show that each factor is continuous then the product is also continuous).