## 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.

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 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!