Spring 2008 #2

Problem Statement: Let $f$ be a twice differentiable real valued function defined on $(a,b)$. Suppose that $a with $f(x_1)>f(x_2)$ and $f(x_3). Prove there exists a $c\in (a,b)$ such that $f''(c)>0$.

Proof: Since $f$ is twice differentiable on $(a,b)$ we know that $f'$ is differentiable on $(a,b)$, which implies that $f'$ is continuous on $(a,b)$.

Since $f$ is differentiable on $(a,b)$ and $(x_1,x_2), (x_2,x_3)\subset (a,b)$ we may apply the Mean Value Theorem on those intervals. The MVT implies that there exists $c_1\in (x_1,x_2)$ and $c_2\in (x_2,x_3)$ such that

$f'(c_1)=\dfrac{f(x_2)-f(x_1)}{x_2-x_1}$

$f'(c_2)=\dfrac{f(x_3)-f(x_2)}{x_3-x_2}$

Note that since $f(x_2) it follows that $f(x_2)-f(x_1)<0$. Since $x_2>x_1$ it follows that $f'(c_1)<0$. Similarly we see that $f(x_3)-f(x_2)>0$ and since $x_3>x_2$ it follows that $f'(c_2)>0$. As stated above $f'$ is continuous on $(a,b)$ and since $(c_1,c_2)\subset (a,b)$ it is continuous on $(c_1,c_2)$. So, by the Intermediate Value Theorem it follows that there is some point $d\in (c_1,c_2)$ such that $f'(d)=0$. Now apply the MVT on the interval $(c_1,d)$. Then there exists some $c\in (c_1,d)$ such that

$f''(c)=\dfrac{f'(d)-f'(c_1)}{d-c_1}=\dfrac{-f'(c_1)}{d-c_1}$

As stated above $f'(c_1)<0$ so $-f'(c_1)>0$. By construction $d>c_1$ and so $d-c_1>0$. Thus $f''(c)=\dfrac{-f'(c_1)}{d-c_1}>0$.

$\Box$

Reflection: When I first approached this problem I knew I had to use the MVT. I got stuck though after finding $f'(c_1), f'(c_2)$. I tried to use MVT again on $(c_1,c_2)$ but I couldn’t guarantee that the difference in the numerator was positive. The key for this problem rested in the fact that differentiable implies continuous. Since we had a continuous derivative function and we knew it was positive and negative at two points, we were guaranteed there was a point in between where the derivative was zero. After there it all fell out nicely.