## Precalculus (6th Edition) Blitzer

The difference quotient for the provided function is $-\frac{1}{x\left( x+h \right)}$.
Consider the provided function: $f\left( x \right)=\frac{1}{x}$. Now, substitute $x=x+h$ in the above equation to find $f\left( x+h \right)$ That is, $f\left( x+h \right)=\frac{1}{x+h}$ Now, apply the difference quotient formula, \begin{align} & \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{\frac{1}{x+h}-\frac{1}{x}}{h} \\ & =\frac{\frac{x-\left( x+h \right)}{x\left( x+h \right)}}{h} \\ & =\frac{-h}{hx\left( x+h \right)} \\ & =-\frac{1}{x\left( x+h \right)} \end{align} Hence, the difference quotient for the provided function is $-\frac{1}{x\left( x+h \right)}$.