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 1


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?)

  • $\begingroup$ yes, thank you so much! $\endgroup$ Aug 20, 2019 at 16:39

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