Let $f$ be a non-negative function which is three-times differentiable on $(a,b)$. If there exist two numbers $c_1$,$c_2$ in $(a,b)$ with $c_1<c_2$ such that $$f(c_1)=f(c_2)=0,$$ proved that there exists $d$ on $(a,b)$ such that $f'''(d)=0.$

I proved $f'(d_1)=0$ ,with $d$ on $(c_1,c_2)$ by Rolle's Theorem. But I don't know how to make things to $f'''(d).$

Some help would be hugely appreciated!

  • $\begingroup$ The statement is false : Let $f$ be cubic function which satisfies $f(c_1)=f(c_2)=0$ and $f(x)>0$ for $x \in (c_1, c_2)$. Then $f'''$ is non-zero constant so there is no $d$ such that $f'''(d)=0$ $\endgroup$
    – bFur4list
    Feb 25 at 6:15
  • 1
    $\begingroup$ The statement is not false; you are forgetting the condition that $f$ be non-negative on the entire open interval $(a,b)$. Indeed there are counterexamples if you only require that $f$ be non-negative on $(c_1, c_2)$. $\endgroup$ Feb 25 at 6:19
  • 2
    $\begingroup$ See math.stackexchange.com/q/692828/42969 $\endgroup$
    – Martin R
    Feb 25 at 6:29

1 Answer 1


Since $f$ is non-negative, $c_1$ and $c_2$ must be local minima of $f$. So, we also have $f''(c_1) \geq 0$ and $f''(c_2) \geq 0$. Moreover, $f$ must attain a maximum on $[c_1, c_2]$. Let $e \in [c_1, c_2]$ be this maximum point.

If $e = c_1$ or $e = c_2$, then the maximum value attained by $f$ on $[c_1, c_2]$ is $0$, so in fact $f(x) = 0$ for all $x \in [c_1, c_2]$. Then we can pick, for example, $d = \frac{c_1 + c_2}{2}$ to get $f'''(d) = 0$.

Otherwise, $e \in (c_1, c_2)$ is a local maximum of $f$, so $f''(e) \leq 0$. By the mean value theorem, there exist points $d_1 \in (c_1, e)$ and $d_2 \in (e,c_2)$ such that

$$f'''(d_1) = \frac{f''(e) - f''(c_1)}{e-c_1} \leq 0$$

$$f'''(d_2) = \frac{f''(c_2) - f''(e)}{c_2 - e} \geq 0.$$

Now by the intermediate value theorem, there must exist some $d \in (d_1,d_2)$ such that $f'''(d) = 0$, as desired.

  • $\begingroup$ You can simplify this. Rolle's Theorem states there exists $e\in (c_1,c_2)$ with $f'(e)=0$. Also $f'(c_1)=0=f'(c_2)$ because $c_1,c_2$ are local minima for $f$. So by the IMT applied to $f',$ there exists $d_1\in (c_1,e)$ and $d_2\in (e,c_2)$ with $f''(d_1)=0=f''(d_2).$ By the IMT applied to $f'',$ there exists $d\in (d_1,d_2)$ with $f'''(d)=0.$ It is easy to forget that in Rolle & in the IMT (which is a direct consequence of Rolle) that the desired point exists strictly between the end-points of the interval. $\endgroup$ Feb 25 at 8:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .