# 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).

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook