The approximate value of sqrt[3]{730} obtaine | vedclass

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(A) Let $f(x) = x^{1/3}$. We want to find the approximate value of $f(730)$.We know that $729 = 9^3$,so let $x = 729$ and $\Delta x = 1$.Using the for

Vedclass · Accepted Answer

$9.0041$

Vedclass · Answer

(A) Let $f(x) = x^{1/3}$. We want to find the approximate value of $f(730)$.We know that $729 = 9^3$,so let $x = 729$ and $\Delta x = 1$.Using the formula for differentials,$f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.Here,$f'(x) = \frac{1}{3} x^{-2/3} = \frac{1}{3x^{2/3}}$.At $x = 729$,$f(729) = (729)^{1/3} = 9$.$f'(729) = \frac{1}{3(729)^{2/3}} = \frac{1}{3(9^2)} = \frac{1}{3 	imes 81} = \frac{1}{243}$.Therefore,$f(730) \approx 9 + \frac{1}{243} 	imes 1$.$f(730) \approx 9 + 0.004115... \approx 9.0041$.

The approximate value of $\sqrt[3]{730}$ obtained by the application of derivatives is

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