How Does the Derivative Change the Multiplicity of a Zero?

[ehrenfest]

Homework Statement

How do you show that if f(t) has a zero t_0 of multiplicity n, then f'(t) has a zero at t_0 of multiplicity EXACTLY n -1?

I can prove that the multiplicity is at least n-1 but I am having trouble with the exactly part...

Homework Equations

The Attempt at a Solution

 

[ircdan]

if f(t) has a root of multiplicity n, then we can write f(t) = (t - t_0)^n g(t) where g(t_0) != 0. Then f'(t) = (t-t_0)^(n-1)k(t) where k(t) = ng(t) + (t-t_0)g'(t) and k(t_0) = ng(t_0) != 0, so f' has a root of multiplicity n-1.