Answer
$$f\left( {x + \Delta x} \right) \approx 2 + \frac{1}{4}\left( {0.02} \right)$$
Work Step by Step
$$\eqalign{
& \sqrt {4.02} \cr
& {\text{Using }}f\left( x \right) = \sqrt x \cr
& {\text{Differentiating}} \cr
& f'\left( x \right) = \frac{1}{{2\sqrt x }} \cr
& {\text{Using the formula}} \cr
& f\left( {x + \Delta x} \right) \approx f\left( x \right) + f'\left( x \right)dx \cr
& {\text{We can write,}} \cr
& f\left( {x + \Delta x} \right) \approx \sqrt x + \frac{1}{{2\sqrt x }}dx \cr
& f\left( {x + \Delta x} \right) = \sqrt {4.02} \cr
& {\text{Now, choosing }}x = 4{\text{ and }}dx = 0.02,{\text{ and substituting }} \cr
& f\left( {x + \Delta x} \right) = \sqrt {4.02} \approx \sqrt 4 + \frac{1}{{2\sqrt 4 }}\left( {0.02} \right) \cr
& f\left( {x + \Delta x} \right) \approx 2 + \frac{1}{4}\left( {0.02} \right) \cr} $$
