Answer
See explanation
Work Step by Step
Approximations for $f(x)=\frac{1}{x}$ at $x_0=2$
{(a)} Let $f(x)=1/x$. Then
\[f'(x)=-\frac{1}{x^2},\qquad f'(2)=-\tfrac{1}{4},\qquad f(2)=\tfrac{1}{2}.
\]
Using Formula (1) the local linear approximation at $x_0=2$ is
\[L(x)=f(2)+f'(2)(x-2)=\tfrac{1}{2}-\tfrac{1}{4}(x-2).
\]
(b)} Writing $\Delta x=x-2$ (Formula (2)) gives
\[L(x)=\tfrac{1}{2}-\tfrac{1}{4}\Delta x.
\]
{(c)} To approximate $dfrac{1}{2.05}$ set $x=2.05$, so $\Delta x=0.05$.
\[L(2.05)=\tfrac{1}{2}-\tfrac{1}{4}(0.05)=0.5-0.0125=0.4875.
\]
For comparison, the exact value is
\[\frac{1}{2.05}\approx 0.4878049.\]
Thus, the linear approximation gives $0.4875$, differing from the exact value by
\[0.4875-0.4878049\approx -0.0003049.\]
Again, the formula in part (b) produces the same numeric result.
