April 2009 #2

Problem Statement: Suppose f is a real-valued differentiable function such that f'(x)\leq 2 for every x\in\mathbb{R}. Show there is at most one u\geq 1 such that f(u)=u^2.

Proof: Suppose to the contrary that there are two such values u_1,u_2\geq 1 such that f(u_1)=(U_1)^2 and f(u_2)=(u_2)^2. Wlog assume that u_1<u_2.

Since f is differentiable on \mathbb{R} we may apply the Mean Value Theorem on the interval [u_1,u_2]. MVT implies that there is some point c\in (u_1,u_2) such that

f'(c)=\dfrac{f(u_2)-f(u_1)}{u_2-u_1}

=\dfrac{(u_2)^2-(u_1)^2}{u_2-u_1}

=\dfrac{(u_2-u_1)(u_2+u_1)}{u_2-u_1}

=u_2+u_1>2

Since 1\leq u_1<u_2 we get the above inequality. But this is a contradiction since we are given that f'(x)\leq 2 for every x\in\mathbb{R}. Thus, there is at most one point u such that f(u)=u^2.

\Box

Reflection: I’m really glad I got this one. When I first saw it I wasn’t exactly sure what was going to happen, but I knew that using MVT would probably be a good idea. Then it all kind of fell out. I’m getting to the point where I know which tools will probably be best to use, even if I don’t necessarily know how the proof is going to play out. This is good I hope!

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

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