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

[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.

 

[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.

 

[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.

 

[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).