Differentials; errors and approximations

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Question

$If the approximate value of (1.999)^{6} is$

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Solution

$\begin{aligned} We write 1.999 = 2 - 0.001 \\ \begin{array}{l} x = 0 and Δ x = - 0.001 \\ f (1.999) - f (2) = 6 (2^{5}) \times (- 0.001) \\ \Rightarrow f (1.999) = 63.81 \end{array} \end{aligned}$

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