Answer
$$\eqalign{
& f\left( {x + \Delta x} \right) \approx 2.962962 \cr
& \root 3 \of {26} \approx 2.96249 \cr} $$
Work Step by Step
$$\eqalign{
& \root 3 \of {26} \cr
& {\text{Using }}f\left( x \right) = \root 3 \of x \cr
& f\left( x \right) = {x^{1/3}} \cr
& {\text{Differentiating}} \cr
& f'\left( x \right) = \frac{1}{3}{x^{ - 2/3}} \cr
& f'\left( x \right) = \frac{1}{{3\root 3 \of {{x^2}} }} \cr
& {\text{Using the formula}} \cr
& f\left( {x + \Delta x} \right) \approx f\left( x \right) + f'\left( x \right)dx \cr
& {\text{We can write,}} \cr
& f\left( {x + \Delta x} \right) \approx \root 3 \of x + \frac{1}{{3\root 3 \of {{x^2}} }}dx \cr
& f\left( {x + \Delta x} \right) = \root 3 \of {26} \cr
& {\text{Now, choosing }}x = 27{\text{ and }}dx = - 1,{\text{ and substituting }} \cr
& f\left( {x + \Delta x} \right) = \root 3 \of {26} \approx \root 3 \of {27} + \frac{1}{{3\root 3 \of {{{\left( {27} \right)}^2}} }}\left( { - 1} \right) \cr
& f\left( {x + \Delta x} \right) \approx 3 - \frac{1}{{27}} \cr
& f\left( {x + \Delta x} \right) \approx 2.962962 \cr
& \cr
& {\text{*Using a calculator to obtain }}\root 3 \of {26} \cr
& \root 3 \of {26} \approx 2.96249 \cr} $$
