The approximate value of sqrt[5]{33} correct | vedclass

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(C) Let $f(x) = x^{1/5}$. We need to find the value of $f(33)$.Let $x = 32$ and $\Delta x = 1$, so that $x + \Delta x = 33$.We know that $f(x) = x^{1/

Vedclass · Accepted Answer

$2.0125$

Vedclass · Answer

(C) Let $f(x) = x^{1/5}$. We need to find the value of $f(33)$.Let $x = 32$ and $\Delta x = 1$, so that $x + \Delta x = 33$.We know that $f(x) = x^{1/5} \implies f'(x) = \frac{1}{5} x^{-4/5} = \frac{1}{5 x^{4/5}}$.At $x = 32$, $f(32) = (32)^{1/5} = 2$.$f'(32) = \frac{1}{5(32)^{4/5}} = \frac{1}{5(2^4)} = \frac{1}{5 	imes 16} = \frac{1}{80} = 0.0125$.Using the differential approximation formula: $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.$f(33) \approx f(32) + f'(32) 	imes 1$.$f(33) \approx 2 + 0.0125 = 2.0125$.

The approximate value of $\sqrt[5]{33}$ correct to $4$ decimal places is

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