Prove that there exists $d$ in$(a,b)$ such that $f'''(d)=0$

Question

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!

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.