## Calculus (3rd Edition)

$$L(x)\approx -\frac{x}{10000}+\frac{1}{50}$$ $$0.01\%~error$$
Given $$\frac{1}{101}$$ Consider $f(x)=\dfrac{1}{x}, a=100,$ and $\Delta x=1$, since \begin{align*} f^{\prime}(x)&=-\frac{1}{x^2} \\ f^{\prime}(100)&=-\frac{1}{10000} \end{align*} Then the linearization to $f (x)$ is given by \begin{align*} L(x)&=f^{\prime}(a)(x-a)+f(a)\\ &=-\frac{1}{10000}(x-100)+\frac{1}{100}\\ &=-\frac{x}{10000}+\frac{1}{50} \end{align*} Since \begin{align*} L(101)&= f(101)\\ &=\frac{1}{101}\\ &\approx-\frac{101}{10000}+\frac{1}{50}\\ &\approx 0.0099 \end{align*} Hence the error is given by $$| \frac{1}{101}-0.0099|=9.900\times 10^{-7}$$ and the percentage is $$\frac{9.900\times 10^{-7} }{1/101}\times 100 \% \approx 0.01\%$$