## May 2000 #3

Problem Statement: Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable when $x=0$. Given that $f(0)=1$ and $f(a+b)=f(a)f(b)$ for every $a,b\in\mathbb{R}$ show that (a)$latex f(x)$ is differentiable on $\mathbb{R}$ and (b) evaluate $f'(x)$.

Proof: First consider $f'(0)=\lim_{h\rightarrow 0}\dfrac{f(0+h)-f(0)}{h}$. We know this limit exists since $f$ is differentiable with $x=0$. Furthermore, since $f(0)=1$ we may conclude that $f'(0)=\lim_{h\rightarrow 0}\dfrac{f(h)-1}{h}=L$

Let $x\in\mathbb{R}$. In order for $f$ to be differentiable at $x$ we must show that $\lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}$ exists.

$f'(x)=\lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}$

$=\lim_{h\rightarrow 0}\dfrac{f(x)f(h)-f(x)}{h}$

$=\lim_{h\rightarrow 0}\dfrac{f(x)(f(h)-1)}{h}$

$=f(x)\lim_{h\rightarrow 0}\dfrac{f(h)-1}{h}$

$=f(x)f'(0)$

Since $f(x)$ is defined for every $x\in\mathbb{R}$ it follows that $f$ is differentiable on $\mathbb{R}-\{0\}$ and that $f'(x)=f(x)f'(1)$ for every $x\in\mathbb{R}-\{0\}$. Since we are given that $f$ is differentiable at $x=0$ it follows that $f$ is differentiable on $\mathbb{R}$

$\Box$

Reflection:  This one wasn’t too bad. The ball dropped when I realized that $f(x)$ was a fixed number and not changing. The main idea here is that since we know $f(a+b)=f(a)f(b)$ we can factor out that $f(x)$ from both terms and end up with a limit we already know about.