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

1/0.502 is very close to 1/0.5 which is 2 (0.5,2) is a point on the graph. Lets find the slope of the tangent point at that point on the graph d/dx (1/x) =-1/x^2 evaluate this at x=0.5 and we get -4

so the tangent line goes through (0.5,2) and has slope -4

this gives an approximation y=-4(x-0.5)+2 Finally let x equal the desired value x=0.502 and evaluate y.