#### Answer

0.003333

#### Work Step by Step

$ f(x)=\sqrt {x+6}$, $ a=3$, $\Delta x=0.02$
In order to find $ f'(x)$, we let $ u=x+6$.
Then, $ f(x)=y=\sqrt u $ and
$ f'(x)=\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=\frac{1}{2\sqrt u}\times1=\frac{1}{2\sqrt {x+6}}$
Recall: $ f(a+\Delta x)-f(a)\approx f'(a)\Delta x $
$\implies f(3+0.02)-f(3)\approx f'(3)\Delta x=\frac{1}{2\sqrt {3+6}}\times0.02=0.003333$