An approximate value of sqrt[4]{18} is | vedclass

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(C) Let $y = f(x) = x^{1/4}$. We choose $x = 16$ and $\Delta x = 2$ because $16$ is the nearest perfect fourth power to $18$. Using the differential f

Vedclass · Accepted Answer

$2.0625$

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(C) Let $y = f(x) = x^{1/4}$. We choose $x = 16$ and $\Delta x = 2$ because $16$ is the nearest perfect fourth power to $18$. Using the differential formula,$\Delta y \approx \left(\frac{dy}{dx}ight) \Delta x$. First,find the derivative: $\frac{dy}{dx} = \frac{1}{4} x^{-3/4} = \frac{1}{4x^{3/4}}$. At $x = 16$,$\frac{dy}{dx} = \frac{1}{4(16^{3/4})} = \frac{1}{4(8)} = \frac{1}{32}$. Now,calculate $\Delta y$: $\Delta y \approx \left(\frac{1}{32}ight) 	imes 2 = \frac{1}{16} = 0.0625$. The approximate value is $y + \Delta y = f(16) + 0.0625 = 2 + 0.0625 = 2.0625$.

An approximate value of $\sqrt[4]{18}$ is

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