## Precalculus (6th Edition) Blitzer

The difference quotient for the provided function is $4x+2h$.
Consider the provided function: $f\left( x \right)=2{{x}^{2}}$. Now, substitute $x=x+h$ in the above equation to find $f\left( x+h \right)$ That is, $f\left( x+h \right)=2{{\left( x+h \right)}^{2}}$ Now, apply the difference quotient formula, \begin{align} & \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{2{{\left( x+h \right)}^{2}}-2{{x}^{2}}}{h} \\ & =\frac{2{{x}^{2}}+4xh+2{{h}^{2}}-2{{x}^{2}}}{h} \\ & =\frac{h\left( 4x+2h \right)}{h} \\ & =4x+2h \end{align} Hence, the difference quotient for the provided function is $4x+2h$.