The approximate value of frac{1}{(2.002)^2} i | vedclass

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(A) Let $f(x) = \frac{1}{x^2} = x^{-2}$.We need to find the value of $f(2.002)$.Let $x = 2$ and $\Delta x = 0.002$.Then $f(x + \Delta x) \approx f(x)

Vedclass · Accepted Answer

$0.2495$

Vedclass · Answer

(A) Let $f(x) = \frac{1}{x^2} = x^{-2}$.We need to find the value of $f(2.002)$.Let $x = 2$ and $\Delta x = 0.002$.Then $f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x$.First,find the derivative $f'(x) = -2x^{-3} = -\frac{2}{x^3}$.At $x = 2$,$f(2) = \frac{1}{2^2} = \frac{1}{4} = 0.25$.And $f'(2) = -\frac{2}{2^3} = -\frac{2}{8} = -0.25$.Now,substitute these values into the approximation formula:$f(2.002) \approx 0.25 + (-0.25) \cdot (0.002)$.$f(2.002) \approx 0.25 - 0.0005 = 0.2495$.Thus,the approximate value is $0.2495$.

The approximate value of $\frac{1}{(2.002)^2}$ is

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