The approximate value of sqrt[3]{28} rounded | vedclass

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Question

(B) Let $f(x) = x^{\frac{1}{3}}$. We need to find the approximate value of $f(28)$.Let $x = 27$ and $\Delta x = 1$, so that $x + \Delta x = 28$.We kno

Vedclass · Accepted Answer

$3.037$

Vedclass · Answer

(B) Let $f(x) = x^{\frac{1}{3}}$. We need to find the approximate value of $f(28)$.Let $x = 27$ and $\Delta x = 1$, so that $x + \Delta x = 28$.We know that $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.Here, $f(x) = x^{\frac{1}{3}}$, so $f'(x) = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}}$.For $x = 27$, $f(27) = (27)^{\frac{1}{3}} = 3$.$f'(27) = \frac{1}{3(27)^{\frac{2}{3}}} = \frac{1}{3(3^3)^{\frac{2}{3}}} = \frac{1}{3(3^2)} = \frac{1}{3 	imes 9} = \frac{1}{27}$.Now, $f(28) \approx f(27) + f'(27) \Delta x$.$f(28) \approx 3 + \left(\frac{1}{27}ight)(1) = 3 + 0.037037...$Rounding to $3$ decimal places, we get $3.037$.Thus, option $(b)$ is correct.

The approximate value of $\sqrt[3]{28}$ rounded up to $3$ decimal places is

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