Answer
$$\eqalign{
& f\left( {x + \Delta x} \right) = 26.73 \cr
& {\left( {2.99} \right)^3} \approx 26.7308 \cr} $$
Work Step by Step
$$\eqalign{
& {\left( {2.99} \right)^3} = {\left( {3 - 0.01} \right)^3} \cr
& {\text{Using }}f\left( x \right) = {x^3} \cr
& {\text{Differentiating}} \cr
& f'\left( x \right) = 3{x^2} \cr
& {\text{Using the formula}} \cr
& f\left( {x + \Delta x} \right) \approx f\left( x \right) + f'\left( x \right)dx \cr
& {\text{We can write}}{\text{,}} \cr
& f\left( {x + \Delta x} \right) \approx {x^3} + 3{x^2}dx{\text{ }}\left( {\bf{1}} \right) \cr
& f\left( {x + \Delta x} \right) = {\left( {2.99} \right)^3} \cr
& {\text{Now}}{\text{, choosing }}x = 3{\text{ and }}dx = - 0.01,{\text{ and substituting into }}\left( {\bf{1}} \right) \cr
& f\left( {x + \Delta x} \right) = {\left( 3 \right)^3} \approx {\left( 3 \right)^3} + 3{\left( 3 \right)^2}\left( { - 0.01} \right) \cr
& f\left( {x + \Delta x} \right) = {\left( 3 \right)^3} + 3{\left( 3 \right)^2}\left( { - 0.01} \right) \cr
& f\left( {x + \Delta x} \right) = 26.73 \cr
& {\left( {2.99} \right)^3} \approx 26.7308 \cr
& \cr
& {\text{*Using a calculator to obtain }}{\left( {2.99} \right)^3} \cr
& {\left( {2.99} \right)^3} \approx 26.7308 \cr} $$
