Find the approximate value of (0.999)^{frac{1 | vedclass

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(A) Let $f(x) = x^{\frac{1}{10}}$.We need to find the value of $f(0.999)$.Let $x = 1$ and $\Delta x = -0.001$,so that $x + \Delta x = 0.999$.Using the

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$0.9999$

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(A) Let $f(x) = x^{\frac{1}{10}}$.We need to find the value of $f(0.999)$.Let $x = 1$ and $\Delta x = -0.001$,so that $x + \Delta x = 0.999$.Using the formula for approximation,$f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.Here,$f(x) = x^{\frac{1}{10}}$,so $f(1) = 1^{\frac{1}{10}} = 1$.The derivative is $f'(x) = \frac{1}{10} x^{\frac{1}{10} - 1} = \frac{1}{10} x^{-\frac{9}{10}} = \frac{1}{10 x^{\frac{9}{10}}}$.At $x = 1$,$f'(1) = \frac{1}{10(1)^{\frac{9}{10}}} = \frac{1}{10} = 0.1$.Now,$f(0.999) \approx f(1) + f'(1) \Delta x$.$f(0.999) \approx 1 + (0.1)(-0.001) = 1 - 0.0001 = 0.9999$.

Find the approximate value of $(0.999)^{\frac{1}{10}}$.

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