Answer
$${\left( {5 + \Delta x} \right)^3} \approx 125 + 75\Delta x$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {4 + x} \right)^3};{\text{ }}{\left( {5 + \Delta x} \right)^3} \approx 125 + 75\Delta x,{\text{ }} \cr
& {x_0} = 1,{\text{ where }}\Delta x = x - 1 \cr
& {\text{Calculate }}f'\left( x \right) \cr
& f'\left( x \right) = 3{\left( {4 + x} \right)^2} \cr
& {\text{Evaluate }}f\left( {{x_0}} \right){\text{ and }}f'\left( {{x_0}} \right){\text{ }} \cr
& f\left( 1 \right) = {\left( {4 + 1} \right)^3} = 125 \cr
& f'\left( 1 \right) = 3{\left( {4 + 1} \right)^2} = 3\left( {25} \right) = 75 \cr
& {\text{Apply }}f\left( {{x_0} + \Delta x} \right) \approx f\left( {{x_0}} \right) + f'\left( {{x_0}} \right)\Delta x{\text{ and substitute the}} \cr
& {\text{ known values, then}} \cr
& {\text{ }}{\left( {5 + \Delta x} \right)^3} \approx 125 + 75\Delta x \cr} $$
