Find the approximate value of (1.999)^{5}. | vedclass

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(A) Let $y = f(x) = x^{5}$.We need to find the value of $(1.999)^{5}$.Let $x = 2$ and $\Delta x = -0.001$,such that $x + \Delta x = 1.999$.Then,$f(x)

Vedclass · Accepted Answer

$31.920$

Vedclass · Answer

(A) Let $y = f(x) = x^{5}$.We need to find the value of $(1.999)^{5}$.Let $x = 2$ and $\Delta x = -0.001$,such that $x + \Delta x = 1.999$.Then,$f(x) = x^{5} = 2^{5} = 32$.The derivative is $f'(x) = 5x^{4}$.Using the formula for approximation,$f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x$.$f'(2) = 5(2)^{4} = 5 	imes 16 = 80$.Therefore,$f(1.999) \approx 32 + 80 	imes (-0.001)$.$f(1.999) \approx 32 - 0.080 = 31.920$.

Find the approximate value of $(1.999)^{5}$.

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