Answer
$$ \approx - 1.016396349$$
Work Step by Step
$$\eqalign{
& \root 3 \of { - 1.05} \cr
& {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found in the exercise 7}} \cr
& f\left( x \right) = \root 3 \of {x - 1} {\text{ we found that}} \cr
& {P_4}\left( x \right) = - 1 + \frac{1}{3}x + \frac{1}{9}{x^2} + \frac{5}{{81}}{x^3} + \frac{{10}}{{243}}{x^4} \cr
& {\text{To approximate }}\sqrt { - 1.05} ,{\text{ we must evaluate }}\sqrt { - 0.05 - 1} ,{\text{ then taking }}x = - 0.05 \cr
& {P_4}\left( { - 0.05} \right) = - 1 + \frac{1}{3}\left( { - 0.05} \right) + \frac{1}{9}{\left( { - 0.05} \right)^2} + \frac{5}{{81}}{\left( { - 0.05} \right)^3} + \frac{{10}}{{243}}{\left( { - 0.05} \right)^4} \cr
& {P_4}\left( { - 0.05} \right) = - 1 - 0.01666666 + 0.00027777 - 0.000007716 + 0.000000257 \cr
& {P_4}\left( { - 0.05} \right) = - 1.016396349 \cr
& {\text{Thus}}{\text{, }}\sqrt { - 1.05} \approx - 1.016396349 \cr} $$
