Calculus problem with Linear Approximation

Question

Use linear approximation, i.e. the tangent line, to approximate  1/0.502 as follows: let f(x)=1/x and find the equation of the tangent line in slope-intercept form to f(x) at a "nice" point near  0.502 . Then use this to approximate 1/0.502.

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Answer 1

Δx = dx = 0.502 - 0.5 = 0.002;

f(0.5) = 1/0.5) = 2;

f'(x) = - 1/x²; f'(0.5) = - 4.

So, f(0.502) ≈ f(0.5) + f'(0.5)·dx = 2 - 4·0.002 = 1.992