You can indeed compute: $$f(3.02)-f(3) = (3.02)^2-3^2 = (6.02)\cdot 0.02 = 0.1204,$$ the exact value. The point of the exercise is avoiding this computations. Using linear approximation gives: $$L(3.02) = f(3) + f'(3)\cdot0.02,$$ so that $\Delta f \approx f'(3)\cdot 0.02 = 6 \cdot 0.02 = 0.12$. Good, no?
Here's how to find the linear approximation:
Let $y=\sqrt{x}$. $\sqrt{3.99} \approx \sqrt{4}=2$. So the difference between $\sqrt{3.99}$ and $2$ should approximately be equal to the differential:
$$dy=y'(x)dx = \left.(\sqrt{x})'\right|_{x=4}(3.99-4)$$
Thus $$\color{red}{\sqrt{3.99} \approx \sqrt{4} + \left.(\sqrt{x})'\right|_{x=4}(3.99-4)}$$
Now let's check our answer. WolframAlpha gives the following values:
$$\sqrt{3.99} = 1.99749843554381789\dots \\ \sqrt{4} + \left.(\sqrt{x})'\right|_{x=4}(3.99-4) = 2+\frac {1}{4}(3.99-4) = 1.9975$$
So you can see that if we get $5$ digits of accuracy (round to the $5$th digit). That's pretty good. Just for fun, though, let's see what the quadratic approximation yields:
The quadratic approximation is $$\color{blue}{\sqrt{3.99} \approx \sqrt{4} + \left.(\sqrt{x})'\right|_{x=4}(3.99-4) +\frac 1{2!}\left.(\sqrt{x})''\right|_{x=4}(3.99-4)^2}$$
Using WolframAlpha again gives:
$$2+ \frac 14(3.99-4) -\frac 1{2!}\frac 1{32}(3.99-4)^2 = 1.9974984375$$
Now we've $9$ digits of accuracy. I wonder how many terms of the Taylor series I need for $100$ digits... :)
Best Answer
Ue Taylor series for $\sqrt{x}$ about $x = 100$. The reason to expand the Taylor series about $100$ is that $100$ is the closest square to $99.2$. $$f(x) = f(100) + f'(100) (x-100) + \text{higher order terms}$$ Hence, $$\sqrt{99.2} \approx \sqrt{100} + \dfrac12 \dfrac{(99.2-100)}{\sqrt{100}} = 10 - \dfrac12 \dfrac{0.8}{10} = 10 - 0.04 = 9.96$$