Linear Approximation

Question

Use linear approximation, i.e. the tangent line, to approximate 1.003 as follows: Let f(x) = ‡ and find the equation of the tangent line to f(x) at a "nice" point near 1.003, f(1.003)). Then use this to approximate 1/1.003.

Answer 1

f(x) = 1/x (I think this is what was meant)

L(x) = f(a)+f'(a)(x-a)

f'(x) = -1/x²

Let a = 1

f(a)=1

f'(a)=-1

L(x)=1-(x-1)

L(1.003) = 0.997