Application of Derivatives rigorous proof Let $f:R\rightarrow R$ be a function such that all its successive derivatives exist in all $R$ and also $f(x)f''(x)\leq 0$ everywhere.
If $\alpha$ and $\beta$ be two successive roots of $f(x)=0$. 
Then prove that $f'''(x)=0$ for atleast one $\gamma \in(\alpha,\beta)$.
My Attempt:
I began by taking an example $f(x)=(x-1)(2-x)$ and let $x=1$ and $x=2$ be its two successive roots.The statement is trivially true.
Then I took the function $f(x)=e^{-x}(x-1)(2-x)$.The statement is true here also so on and so forth.
But what would be the exact proof I wonder. It appears to be a question of Rolle's Theorem.
 A: $f(x)f''(x)< 0$ for all $x$ is problematic if $f(\alpha) = 0.$
If we changed this to:
$f(x)f''(x)\le 0$ you are in business.
What you need to do is show that $\alpha, \beta$ must also be roots of $f''(x).$  Since f is smooth, and the constraint above, that shouldn't be too hard.
and then by Rolles theorem, there exist a $\gamma \in (\alpha,\beta)$ where $f'''(\gamma) = 0$ 
A: If $f(x)>0$ for $x$ between two successive roots of $f(x)=0$, then $f''(x)<0$ for all values of $x$ between those roots.  That implies there is some minimum point $c$ between those two roots, so that $f''(c)\le f''(x)$ for all $x$ between those roots.  At a local minimum of $f''$, the first derivative of $f''$ must be $0$.  So at that point you have $f'''(c)=0$.
A: Since $\alpha,\beta$ are successive roots of $f(x)$ it follows that $f(x)$ is of constant sign in $(\alpha, \beta)$. Let's assume that $f$ is positive on interval (and the proof for the case when $f$ is negative is similar). By Rolle's Theorem $f'$ vanishes at least once in the interval $(\alpha, \beta)$.
Now we assume on the contrary that $f'''(x)$ never vanishes in $(\alpha, \beta)$ so that it is of constant sign (via Darboux Theorem). Hence $f''$ is strictly monotone in $[\alpha, \beta]$. Moreover since $f$ is positive and $f(x)f''(x)\leq 0$ it follows that $f''$ must be zero or negative in $(\alpha, \beta)$. Let's assume that $f''$ is strictly increasing and hence in that case $f''$ is negative on $[\alpha, \beta)$. And by continuity of $f''$ it is negative in a neighborhood of $\alpha$. This means that $f'$ is strictly decreasing in this neighborhood of $\alpha$ as well as in $[\alpha, \beta]$. Since $f'$ vanishes once in $(\alpha, \beta)$ it follows that $f'$ is positive at $\alpha$ and in its neighborhood. Therefore $f$ is strictly increasing in neighborhood of $\alpha$ and since $f(\alpha) = 0$ it means that $f$ changes sign as it crosses $\alpha$. This is not possible because $f''$ maintains a constant sign in neighborhood of $\alpha$ and $f(x)f''(x) \leq 0$. This gives us the desired contradiction.
Note that if $f''(x)$ were strictly decreasing then all the above action would be happening at $\beta$ and in its neighborhood to arrive at a similar contradiction.

Another approach which avoids the use of Darboux theorem is as follows. As before let $f$ be positive in $(\alpha, \beta)$ and $f(\alpha) = f(\beta) = 0$. Consider the point $\alpha$. If $f''(\alpha) \neq 0$ then $f''(x)$ maintains constant sign in a neighborhood of $\alpha$ and clearly since $f(x)f''(x) \leq 0$ it follows that $f''$ is negative in this neighborhood of $\alpha$. From the equation $f(x)f''(x) \leq 0$ it follows that $f(x) \geq 0$ in this neighborhood of $\alpha$ and therefore $\alpha$ is a local minimum of $f$ and hence $f'(\alpha) = 0$. But $f'(\alpha) = 0$ and $f''(\alpha) < 0$ imply that $\alpha$ is a local strict maximum of $f$ and this contradiction shows that $f''(\alpha) = 0$. By applying same argument at $\beta$ we get $f''(\beta) = 0$ hence by Rolle's Theorem $f'''(\gamma) = 0$ for some $\gamma \in (\alpha, \beta)$.
