Find the approximate value of (255)^{frac{1}{ | vedclass

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(A) Let $f(x) = x^{\frac{1}{4}}$.We know that $256^{\frac{1}{4}} = 4$.Let $x = 256$ and $\Delta x = -1$.Then,$f(x + \Delta x) = (x + \Delta x)^{\frac{

Vedclass · Accepted Answer

$3.9961$

Vedclass · Answer

(A) Let $f(x) = x^{\frac{1}{4}}$.We know that $256^{\frac{1}{4}} = 4$.Let $x = 256$ and $\Delta x = -1$.Then,$f(x + \Delta x) = (x + \Delta x)^{\frac{1}{4}} = (255)^{\frac{1}{4}}$.The approximate value is given by $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.Here,$f'(x) = \frac{1}{4} x^{-\frac{3}{4}} = \frac{1}{4 x^{\frac{3}{4}}}$.For $x = 256$,$f'(256) = \frac{1}{4 (256)^{\frac{3}{4}}} = \frac{1}{4 (4^4)^{\frac{3}{4}}} = \frac{1}{4 	imes 4^3} = \frac{1}{4 	imes 64} = \frac{1}{256} = 0.00390625$.Thus,$f(255) \approx f(256) + f'(256) \Delta x = 4 + (0.00390625)(-1) = 4 - 0.00390625 = 3.99609375$.Rounding to four decimal places,we get $3.9961$.

Find the approximate value of $(255)^{\frac{1}{4}}$.

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