#### Answer

$$ \approx 4.037325847$$

#### Work Step by Step

$$\eqalign{
& \sqrt {16.3} \cr
& {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found in the exercise 6}} \cr
& f\left( x \right) = \sqrt {x + 16} {\text{ we found that}} \cr
& {P_4}\left( x \right) = 4 + \frac{1}{8}x - \frac{1}{{512}}{x^2} + \frac{1}{{16,384}}{x^3} - \frac{5}{{2,097,152}}{x^4} \cr
& {\text{To approximate }}\sqrt {16.3} ,{\text{ we must evaluate }}\sqrt {0.3 + 16} ,{\text{ then taking }}x = 0.3 \cr
& {P_4}\left( {0.3} \right) = 4 + \frac{1}{8}\left( {0.3} \right) - \frac{1}{{512}}{\left( {0.3} \right)^2} + \frac{1}{{16,384}}{\left( {0.3} \right)^3} - \frac{5}{{2,097,152}}{\left( {0.3} \right)^4} \cr
& {P_4}\left( {0.3} \right) = 4 + 0.0375 - 0.000175781 + 0.000001647 - 0.000000019 \cr
& {P_4}\left( {0.3} \right) = 4.037325847 \cr
& {\text{Thus}}{\text{, }}\sqrt {16.3} \approx 4.037325847 \cr} $$