Proving the existence of an inflection point. If $f(x)$ meets the conditions of the Mean Value Theorem on $[a,b]$ and has more than one value $c$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$ , then does there exist at least one inflection point on $(a,b)$? It seems intuitively to be true but I'm not sure how to prove it.
 A: Assuming that $f$ is twice differentiable (otherwise the notion of point of inflection is undefined), it is easy to show that the second derivative must be zero somehere: just apply MVT to $f'(x)$ in the interval between the two values $c_1$ and $c_2$, to get $\xi\in(c_1,c_2)$ such that $f''(\xi)=(f'(c_2)-f'(c_1))/(c_2-c_1)=0$.
But as David Mitra points out in a comment, that is not enough to guarantee the existence of a point of inflection; the second derivative must also change sign. As David says, any linear function serves as a counterexample. In fact, the function only has to be linear on the interval $[c_1,c_2]$ to serve as a counterexample.
If $f$ is not linear on $[c_1,c_2]$, then there exists $\eta\in(c_1,c_2)$ such that $f'(\eta)\ne f'(c_1)$. Suppose without loss of generality that $f'(\eta)>f'(c_1)$. Then (by the Mean Value Theorem applied to $f'$ on $[c_1,\eta]$) there exists $\psi\in(c_1,\eta)$ such that $f''(\psi)>0$; and similarly, there exists $\phi\in(\eta,c_2)$ such that $f''(\phi)<0$. So if $f''$ is continuous on $(c_1,c_2)$, it must change sign somewhere in $(c_1,c_2)$, and we have our point of inflection. (If $f''$ is not continuous, then $f$ doesn't have a well-defined curvature everywhere, so the notion of point of inflection is not well-defined.)
