Find the approximate value of (26)^{frac{1}{3 | vedclass

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Question

(B) Let $f(x) = x^{\frac{1}{3}}$.We know that $27^{\frac{1}{3}} = 3$.Let $x = 27$ and $\Delta x = -1$.Then $f(x + \Delta x) = (x + \Delta x)^{\frac{1}

Vedclass · Accepted Answer

$2.9630$

Vedclass · Answer

(B) Let $f(x) = x^{\frac{1}{3}}$.We know that $27^{\frac{1}{3}} = 3$.Let $x = 27$ and $\Delta x = -1$.Then $f(x + \Delta x) = (x + \Delta x)^{\frac{1}{3}} = (26)^{\frac{1}{3}}$.The approximate value is given by $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.Here,$f'(x) = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3(x^{\frac{1}{3}})^2}$.For $x = 27$,$f'(27) = \frac{1}{3(27^{\frac{1}{3}})^2} = \frac{1}{3(3)^2} = \frac{1}{27}$.Thus,$(26)^{\frac{1}{3}} \approx 3 + (\frac{1}{27})(-1) = 3 - 0.037037... \approx 2.96296...$Rounding to four decimal places,we get $2.9630$.

Find the approximate value of $(26)^{\frac{1}{3}}$.

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