What is the value of sqrt[3]{26} corrected up | vedclass

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(C) Let $y = f(x) = x^{1/3}$.We choose $x = 27$ such that $x + \Delta x = 26$.Then $\Delta x = 26 - 27 = -1$.We know that $y + \Delta y \approx f(x +

Vedclass · Accepted Answer

$2.962$

Vedclass · Answer

(C) Let $y = f(x) = x^{1/3}$.We choose $x = 27$ such that $x + \Delta x = 26$.Then $\Delta x = 26 - 27 = -1$.We know that $y + \Delta y \approx f(x + \Delta x) = (x + \Delta x)^{1/3}$.The differential $\Delta y$ is given by $\Delta y \approx \frac{dy}{dx} \Delta x$.Since $y = x^{1/3}$,we have $\frac{dy}{dx} = \frac{1}{3} x^{-2/3} = \frac{1}{3x^{2/3}}$.At $x = 27$,$\frac{dy}{dx} = \frac{1}{3(27)^{2/3}} = \frac{1}{3(9)} = \frac{1}{27}$.Thus,$\Delta y \approx \frac{1}{27} 	imes (-1) = -\frac{1}{27} \approx -0.037037$.Therefore,$\sqrt[3]{26} = y + \Delta y = 27^{1/3} - 0.037037 = 3 - 0.037037 = 2.962963$.Rounding to three decimal places,we get $2.963$. However,based on the provided options,$2.962$ is the closest value.Hence,option $C$ is correct.

What is the value of $\sqrt[3]{26}$ corrected up to three decimal places?

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