Answer
A) The slope of the tangent to the graph of $f\left( x \right)=\frac{1}{x}$ at $\left( 1,1 \right)$ is $-1$.
B) The slope intercept equation of the tangent line is $y=-x+2$.
Work Step by Step
(a)
Consider the function $f\left( x \right)=\frac{1}{x}$ and the point $\left( 1,1 \right)$
Here, $\left( a,f\left( a \right) \right)=\left( 1,1 \right)$
Now, compute the slope of the tangent line for the function $f\left( x \right)=\frac{1}{x}$ as follows:
Put $a=1$ ,
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 1+h \right)-f\left( 1 \right)}{h}$
To compute $f\left( 1+h \right)$, substitute $x=1+h$ in the function $f\left( x \right)=\frac{1}{x}$.
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{1}{1+h}-\frac{1}{1}}{h}$
Now, simplify the above expression by taking the LCM of the denominator as follows:
$\begin{align}
& {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{1-\left( 1+h \right)}{1+h}}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{1-1-h}{h\left( 1+h \right)}
\end{align}$
Combine the like terms in the numerator and divide the numerator as well as the denominator by h,
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{-1}{1+h}$
Apply the limits and simplify.
$\begin{align}
& {{m}_{\tan }}=\frac{-1}{1+0} \\
& =-1
\end{align}$
(b)
Consider the function $f\left( x \right)=\frac{1}{x}$ and the point $\left( 1,1 \right)$.
From part (a), the slope of the tangent to the graph of $f\left( x \right)=\frac{1}{x}$ at $\left( 1,1 \right)$ is $-1$.
Now compute the slope intercept equation of the tangent line as follows:
Here, the tangent line passes through the point $\left( 1,1 \right)$ which implies ${{x}_{1}}=1$ and ${{y}_{1}}=1$
Substitute $m=-1,{{x}_{1}}=1,{{y}_{1}}=1$ in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.
$\begin{align}
& y-1=-1\left( x-1 \right) \\
& y-1=-x+1 \\
& y=-x+1+1 \\
& y=-x+2
\end{align}$
