The approximate value of (66)^{frac{1}{3}} is | vedclass

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(A) Let $f(x) = x^{\frac{1}{3}}$. We want to find the approximate value of $f(66)$.Let $x = 64$ and $\Delta x = 2$,so that $x + \Delta x = 66$.The for

Vedclass · Accepted Answer

$4.0416$

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(A) Let $f(x) = x^{\frac{1}{3}}$. We want to find the approximate value of $f(66)$.Let $x = 64$ and $\Delta x = 2$,so that $x + \Delta x = 66$.The formula for linear approximation is $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.Here,$f(x) = x^{\frac{1}{3}}$,so $f'(x) = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3 x^{\frac{2}{3}}}$.At $x = 64$,$f(64) = (64)^{\frac{1}{3}} = 4$.And $f'(64) = \frac{1}{3(64)^{\frac{2}{3}}} = \frac{1}{3(4^2)} = \frac{1}{3(16)} = \frac{1}{48}$.Now,$f(66) \approx f(64) + f'(64) \Delta x$.$f(66) \approx 4 + \left(\frac{1}{48}\right)(2) = 4 + \frac{1}{24}$.Since $\frac{1}{24} \approx 0.04166...$,we have $f(66) \approx 4 + 0.04166... = 4.04166...$.Rounding to four decimal places,we get $4.0417$,which is closest to $4.0416$.

The approximate value of $(66)^{\frac{1}{3}}$ is

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