Answer
$$-6.25\times 10^{-3},\ \ \ -6.13520\times 10^{-3} ,\ \ \ 0.0001148$$
Work Step by Step
Given $$\frac{1}{\sqrt{4.1}}-\frac{1}{2} $$
Consider $f(x)=x^{-1 / 2}, a=4,$ and $\Delta x=0.1$, since
\begin{align*}
f^{\prime}(x)&=\frac{-1}{2}x^{-3/2}\\
f^{\prime}(4)&=\frac{-1}{16}
\end{align*}
Then
\begin{align*}
\Delta f&=f (a+\Delta x)-f(a)\\
&\approx f'(a)\Delta x\\
&\approx \frac{-0.1}{16} \\
&\approx -6.25\times 10^{-3}
\end{align*}
By using a calculator, $\frac{1}{\sqrt{4.1}}-\frac{1}{2}=-6.13520\times 10^{-3}$ and the error in the linear approximation is$$
|6.13520\times 10^{-3}-6.25\times 10^{-3}|=0.0001148
$$
