Exercise 8.6

(a) By assumption, there is a real $x$ such that $f(x)\ne 0$. Hence $f(x)=f(x+0)=f(x)f(0)$, showing that $f(0)=1$. Since $f$ is differentiable, we have for all real $x$

So, letting $c=f^{\prime }(0)$, we have $f^{\prime }(x)=c{f}^{\prime }(x)$ for all real $x$. Since this is also true for the function $g(x)=e^{\mathit{cx}}$, we have

so that $f=\mathit{Kg}$ for some constant $K$. Since $f(0)=1=g(0)$, we have $K=1$, or $f(x)=g(x)=e^{\mathit{cx}}$.

(b) For all rational numbers $p=m∕n$, we have $f(p)=f{(1)}^{m∕n}$. Let $c=\log <!--\; FUNCTION\; APPLICATION\; -->f(1)$, and let $g(x)=e^{\mathit{cx}}$. Since we also have $g(p)=g{(1)}^{m∕n}$ and $g(1)=e^{\log <!--\; FUNCTION\; APPLICATION\; -->f(1)}=f(1)$, we have $f(x)=g(x)$ on all rational numbers. Hence, since the continuous functions $f$ and $g$ are equal on a dense subset of the real numbers, they are equal for all real numbers.