Answer
$$ y=-\frac{1}{16} x+\frac{1}{2}$$
Work Step by Step
Given $$f(x)=x^{-1}, \quad a=4$$
Since $f(4)=0.25$ and
$$ f^{\prime}(a) =\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$$
Then
\begin{aligned}
f^{\prime}(4) & =\lim _{h \rightarrow 0} \frac{\frac{1}{4+h} - \frac{1}{4}}{h} \\
&=\lim _{h \rightarrow 0} \frac{4-(4+h)}{4 h(4+h)}\\
&=\lim _{h \rightarrow 0} \frac{-1}{4(4+h)}\\
&=-\frac{1}{16}
\end{aligned}
Then the tangent line at $x=4$ is given by
\begin{align*}
\frac{y-y_1}{x-x_1}&=f'(a)\\
\frac{y-0.25}{x-4}&=\frac{-1}{16}\\
y-\frac{1}{4}&= \frac{-1}{16}(x-4)\\
\end{align*}
Hence $$ y=-\frac{1}{16} x+\frac{1}{2}$$
