## Calculus (3rd Edition)

$$0.98,\ \ 0.980392,\ \ 3.922\times 10^{-4}$$
Given $$\frac{1}{1.02}$$ Consider $f(x)=\frac{1}{x}, a=1,$ and $\Delta x=0.02$, since \begin{align*} f^{\prime}(x)&=\frac{-1}{x^2} \\ f^{\prime}(1)&=-1 \end{align*} Then the linear approximation is given by \begin{align*} L(x)&=f^{\prime}(a)(x-a)+f(a)\\ &= -(x-1)+ 1=2-x\\ L(1.02)&\approx 2-1.02= 0.98 \end{align*} By using a calculator, $\frac{1}{1.02} =0.980392$ and the error in the linear approximation is$$|0.980392- 0.98|=3.922\times 10^{-4}$$