Question

Use linear approximation, i.e. the tangent line, to approximate 1/0.503 as follows: Let f(x)=1/x and find the equation of the tangent line to f(x) at a "nice" point near 0.503. Then use this to approximate 1/0.503.

Answer 1

f'(x) = -1/x^2
the nice number would be 0.5, or 1/2

Answer 2

f'(x)=-1/x^2 so this gives you the slope of your tangent line. The nice point is .5