## Calculus: Early Transcendentals 8th Edition

$f(x)=e^x$ and $a=-2$
*According to definition, the derivative of a function $f$ at a number $a$ is $$f'(a)=\lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}$$ Here we have $$f'(a)=\lim\limits_{h\to0}\frac{e^{-2+h}-e^{-2}}{h}$$ Now we match the formula found above with the formula of the derivative according to definition. We find that $f(a+h)=e^{-2+h}$ and $f(a)=e^{-2}$ So, $a=-2$ and $f(a)=f(-2)=e^{-2}$ Therefore, $f(x)=e^{x}$