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 to f(x) at a "nice" point near 0.502. Then use this to approximate 1/0.502.

Answer 1

df(x)/dx=-1/x^2
at x=.5 dx=.02
df(x)=-.08
y=1/.5=2
1/.502=2-.08=1.92 approx.