Answer
$$ \approx 2.986636909$$
Work Step by Step
$$\eqalign{
& \sqrt {8.92} \cr
& {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found in the exercise 5}} \cr
& f\left( x \right) = \sqrt {x + 9} {\text{ we found that}} \cr
& {P_4}\left( x \right) = 3 + \frac{1}{6}x - \frac{1}{{216}}{x^2} + \frac{1}{{3888}}{x^3} - \frac{5}{{279,936}}{x^4} \cr
& {\text{To approximate }}\sqrt {8.92} ,{\text{ we must evaluate }}\sqrt { - 0.08 + 9} ,{\text{ then taking }}x = - 0.08 \cr
& {P_4}\left( {8.92} \right) = 3 + \frac{1}{6}\left( { - 0.08} \right) - \frac{1}{{216}}{\left( { - 0.08} \right)^2} + \frac{1}{{3888}}{\left( { - 0.08} \right)^3} - \frac{5}{{279,936}}{\left( { - 0.08} \right)^4} \cr
& {P_4}\left( {8.92} \right) = 3 - 0.013333333 - 0.000029629 - 0.000000131 - 0.000000731 \times {10^{ - 3}} \cr
& {P_4}\left( {8.92} \right) = 2.986636909 \cr
& {\text{Thus}}{\text{, }}\sqrt {8.92} \approx 2.986636909 \cr} $$
