# Thread: Function having a third derivative

1. ## Function having a third derivative

Let $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ have a third derivative that exists at all points. Show that there must exist at least one point $\displaystyle \xi$ for which
$\displaystyle f(\xi)f^{'}(\xi)f^{''}(\xi)f^{'''}(\xi) \geq 0$

2. Originally Posted by Markeur
Let $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ have a third derivative that exists at all points. Show that there must exist at least one point $\displaystyle \xi$ for which
$\displaystyle f(\xi)f^{'}(\xi)f^{''}(\xi)f^{'''}(\xi) \geq 0$
Once again, any ideas?

3. Start by using Mean Value Theorem?How do I start from there?

4. Originally Posted by Markeur
Start by using Mean Value Theorem?How do I start from there?

I'd use the MVT, but how?

5. Hmm

Restrict the function to $\displaystyle f:[a,b] \rightarrow \mathbb{R}$, where $\displaystyle f, f^{'}, f^{''}$ are increasing in that interval and $\displaystyle f(x) \geq 0$ for $\displaystyle x \in [a,b]$. Since $\displaystyle f$ is continuous, by MVT, there exists $\displaystyle c \in [a,b]$ such that $\displaystyle \frac{f(b)-f(a)}{b-a} = f^{'}(c)$. Then $\displaystyle f^{'}(c) \geq 0$. Since $\displaystyle f^{'}$ is continuous, by MVT, there exists $\displaystyle d \in [c,b]$ such that $\displaystyle \frac{f^{'}(b)-f^{'}(c)}{b-c} = f^{''}(d)$. Then $\displaystyle f^{''}(d) \geq 0$. Since $\displaystyle f^{''}$ is continuous, by MVT, there exists $\displaystyle \xi \in [d,b]$ such that $\displaystyle \frac{f^{''}(b)-f^{''}(d)}{b-d} = f^{'''}(\xi)$. Then $\displaystyle f^{'''}(\xi) \geq 0$. Since $\displaystyle f, f^{'}, f^{''}$ are increasing in that interval, therefore $\displaystyle f^{'}(\xi) \geq f^{'}(c)$ and $\displaystyle f^{''}(\xi) \geq f^{''}(d)$. These imply that $\displaystyle f(\xi), f^{'}(\xi), f^{''}(\xi), f^{'''}(\xi) \geq 0$. Hence the conclusion holds.

This could be one possible case out of so many different cases. Anyway, is this particular proof correct?

First, if any one of the four functions $\displaystyle f,\ f',\ f'',\ f'''$ vanishes at some point $\displaystyle \xi$, then $\displaystyle f(\xi)f'(\xi)f''(\xi)f'''(\xi) = 0$ and so the result is proved. It follows from the intermediate value theorem that each of the (continuous) functions $\displaystyle f,\ f',\ f''$ always has the same sign. By Darboux's theorem the same is true for $\displaystyle f'''$.
Replacing $\displaystyle f$ by $\displaystyle -f$ if necessary, you can assume that $\displaystyle f$ is always positive
Now suppose that $\displaystyle f'$ is always negative. By considering what happens as $\displaystyle \xi\to\infty$, show that $\displaystyle f''$ must be positive. The same argument, repeated, will show that $\displaystyle f'''$ is negative. Therefore the product $\displaystyle ff'f''f'''$ must be positive.
By considering what happens as $\displaystyle \xi\to-\infty$, use the same ideas to show that if $\displaystyle f'$ is positive then all four functions must be positive, hence so is their product.