Answer
$$ \approx 1.0149991075$$
Work Step by Step
$$\eqalign{
& \root 4 \of {1.06} \cr
& {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found in the exercise 9}} \cr
& f\left( x \right) = \root 4 \of {x + 1} {\text{ we found that}} \cr
& {P_4}\left( x \right) = 1 + \frac{1}{4}x - \frac{3}{{32}}{x^2} + \frac{7}{{128}}{x^3} - \frac{{77}}{{2048}}{x^4} \cr
& {\text{To approximate }}\root 4 \of {1.06} ,{\text{ we must evaluate }}\sqrt {0.06 + 1} ,{\text{ then taking }}x = 0.06 \cr
& {P_4}\left( {0.06} \right) = 1 + \frac{1}{4}\left( {0.06} \right) - \frac{3}{{32}}{\left( {0.06} \right)^2} + \frac{7}{{128}}{\left( {0.06} \right)^3} - \frac{{77}}{{2048}}{\left( {0.06} \right)^4} \cr
& {P_4}\left( {0.06} \right) = 1 + 0.015 - 0.00002025 + 0.000011812 - 0.000000487 \cr
& {P_4}\left( {0.06} \right) = 1.0149991075 \cr
& {\text{Thus}}{\text{, }}\root 4 \of {1.06} \approx 1.0149991075 \cr} $$
