## Precalculus (6th Edition) Blitzer

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 $y=-x+2$.
(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}