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

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 $$c_1 such that $$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!

• 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 $f'''(d)=0$ Feb 25 at 6:15
• 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 $(c_1, c_2)$. Feb 25 at 6:19
• Feb 25 at 6:29

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.

• 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. Feb 25 at 8:39