## Calculus (3rd Edition)

Given/Known: $f(x)=x^{-1}$, $a=3$, $\Delta x=0.02$ and $f'(x)=-\frac{1}{x^{2}}$ Recall: $f(a+\Delta x)-f(a)=f'(a)\Delta x$ $\implies$ $f(3+0.02)-f(3)=f'(3)\times\Delta x=-\frac{1}{3^{2}}\times0.02$ $=-0.00222$