# Prove that if $f$ is continuous and differentiable on $\mathbb{R}$ and has three roots, then its derivative $f′$ has at least two roots.

I know that we are supposed to use Mean Value THeorem for this question. So from the theorem, if $$f$$ is continuous on an interval $$[a,b]$$ and has two roots, this means that there is a point $$c\in [a,b]$$ where $$f'(x)=0.$$ But is this logic correct for $$3$$ roots?

I am not sure if I understand how to construct a proof here.

## 1 Answer

If $$f$$ is differentiable and $$f(x_1) = f(x_2) = f(x_3) = 0$$ with $$x_1 < x_2 < x_3$$ then you can apply the mean-value theorem (or Rolle's theorem) to both intervals $$[x_1,x_2]$$ and $$[x_2, x_3]$$. It follows that $$f'$$ has a root in each of the open intervals $$(x_1, x_2)$$ and $$(x_2, x_3)$$. That makes (at least) two roots of the derivative.

In the same way you can show that if $$f$$ has $$n$$ distinct roots then $$f'$$ has (at least) $$n-1$$ distinct roots.

(And if the second derivative exists, then $$f''$$ has at least $$n-2$$ roots, you get the idea?)

• yes, thank you so much! Aug 20, 2019 at 16:39