Spring 2008 #2

Problem Statement: Let f be a twice differentiable real valued function defined on (a,b). Suppose that a<x_1<x_2<x_3<b with f(x_1)>f(x_2) and f(x_3)<f(x_2). 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)<f(x_1) 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.

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This entry was posted in Analysis, Continuity, Differentiable, Math, MVT. Bookmark the permalink.

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