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.

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