## Winter 1995 #3

Problem Statement: Let $f:[0,1]\rightarrow\mathbb{R}$ be continuous, $f(0)=0$, and $f'(x)$ exists for every $x\in (0,1)$. Let $f'(x)$ be increasing on $(0,1)$. Show $g(x)=\dfrac{f(x)}{x}$ is increasing on $(0,1)$.

Proof: We wish to show that $g(x)$ is increasing on $(0,1)$ which means that $g'(x)>0$ for $x\in(0,1)$.

Since we know that $f$ is continuous and differentiable we may use the Mean Value Theorem. Let $x\in (0,1)$, then the MVT implies that there is some $c\in (0,x)$ such that $f'(c)=\dfrac{f(x)-f(0)}{x-o}=\dfrac{f(x)}{x}=g(x)$. Consider $g'(x)=f''(c)$. We are given that $f'$ is increasing on $(0,1)$, this implies that $f''(c)>0$ for all $c\in (0,1)$. Thus, $g'(x)=f''(c)>0$. Since $x$ was arbitrary in $(0,1)$ and $f''(c)>0$ for all $c\in (0,1)$ this implies that $g'(x)>0$ for every $x\in (0,1)$.

Thus, $g(x)$ is increasing on $(0,1)$.

$\Box$

Reflection: When I first approached this problem I think I was over-thinking it. I got the equality I have above, but I was worried about concluding that $g'(x)>0$ for all $x\in (0,1)$ since I only showed it was equal to some $f'(c)$. The thing that saves us is that our $x$ was arbitrary, so we can define $g(x)$ for every $x\in (0,1)$ and show that no matter which $x$ we’re looking at $g'(x)>0$. The only reason we’re able to conclude anything in this proof is because we know $f'(x)$ is increasing. Without that we wouldn’t be able to say anything.