Calculus 8th Edition

$f(x) = \sqrt x$ $a = 9$
The form of the equation given resembles the form $\lim\limits_{h \to 0}$ $\frac{f(a+h)-f(a)}{h}$ In this case, we have $f(a+h)$ = $\sqrt {9+h}$, so $a$ must be 9. This is further supported by the $f(a)$ portion, which shows that $f(9)$ = 3, so our function $f(x)$ must be $\sqrt x$