Answer
4.02
Work Step by Step
Take y= $x^{\frac{1}{3}}$.
Let x= 64 and let Δx= 1.
Then, Δy=$ (x+Δx)^{\frac{1}{3}}-x^{\frac{1}{3}}=(65)^{\frac{1}{3}}-(64)^{\frac{1}{3}}$
= $(65)^{\frac{1}{3}}-4$ or $(65)^{\frac{1}{3}}= 4+Δy.$
As dx=Δx is relatively small when compared with x, dy is approximately equal to Δy and is given by, dy= $(\frac{dy}{dx})Δx=\frac{1}{3x^{\frac{2}{3}}}\times1$
= $\frac{1}{3((64)^{\frac{1}{3}})^{2}}=\frac{1}{48}$.
Thus the approximate value of $\sqrt[3] 65$ is $4+\frac{1}{48}= \frac{193}{48}\approx 4.02$
