## Precalculus (6th Edition) Blitzer

a) The derivative of $f\left( x \right)=\frac{8}{x}$ at x is $f'\left( x \right)=-\frac{8}{{{x}^{2}}}$. b) The slope of the tangent line to the graph of $f\left( x \right)=\frac{8}{x}$ at $x=-2$ is $f'\left( -2 \right)=-2$ and at $x=1$ is $f'\left( 1 \right)=-8$.
(a) Consider the function, $f\left( x \right)=\frac{8}{x}$ Now, compute the derivate of $f\left( x \right)=\frac{8}{x}$ using the formula $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$ To compute $f\left( x+h \right)$, substitute $x=x+h$ in the function $f\left( x \right)=\frac{8}{x}$. $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{8}{x+h}-\frac{8}{x}}{h}$ Now, simplify the above expression by taking the LCM of the denominator as follows: \begin{align} & f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{8x-8\left( x+h \right)}{x\left( x+h \right)}}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{8x-8x-8h}{hx\left( x+h \right)} \end{align} Combine the like terms in the numerator, then divide numerator and denominator by h. $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{-8}{x\left( x+h \right)}$ Apply the limits and simplify, \begin{align} & f'\left( x \right)=\frac{-8}{x\left( x+0 \right)} \\ & =-\frac{8}{{{x}^{2}}} \end{align} Thus, the derivative of $f\left( x \right)=\frac{8}{x}$ at x is $f'\left( x \right)=-\frac{8}{{{x}^{2}}}$. (b) Consider the function, $f\left( x \right)=\frac{8}{x}$ From part (a), the derivative of $f\left( x \right)=\frac{8}{x}$ at x is $f'\left( x \right)=-\frac{8}{{{x}^{2}}}$. To compute the slope of the tangent line at $x=-2$, substitute $x=-2$ in $f'\left( x \right)$. \begin{align} & f'\left( -2 \right)=-\frac{8}{{{\left( -2 \right)}^{2}}} \\ & =-\frac{8}{4} \\ & =-2 \end{align} To compute the slope of the tangent line at $x=1$, substitute $x=1$ in $f'\left( x \right)$. \begin{align} & f'\left( 1 \right)=-\frac{8}{{{\left( 1 \right)}^{2}}} \\ & =-\frac{8}{1} \\ & =-8 \end{align}