## Precalculus (6th Edition) Blitzer

The required solution is $8x+4h-5$.
Replace $f\left( x \right)$ with $f\left( x+h \right)$, that is., ${{x}^{2}}={{\left( x+h \right)}^{2}}\text{and }x=x+h$ Then, \begin{align} & f\left( x+h \right)=4{{(x+h)}^{2}}-5\left( x+h \right)-2 \\ & f\left( x+h \right)=4\left( {{x}^{2}}+2xh+{{h}^{2}} \right)-5x-5h-2 \\ & f\left( x+h \right)=4{{x}^{2}}+8xh+4{{h}^{2}}-5x-5h-2 \\ \end{align} Substitute the values of $f\left( x+h \right)$ in the provided equation. That is, \begin{align} & f\left( x \right)=\frac{f\left( x+h \right)-f\left( x \right)}{h} \\ & f\left( x \right)=\frac{4{{x}^{2}}+8xh+4{{h}^{2}}-5x-5h-2-\left( 4{{x}^{2}}-5x-2 \right)}{h} \\ & f\left( x \right)=\frac{4{{x}^{2}}+8xh+4{{h}^{2}}-5x-5h-2-4{{x}^{2}}+5x+2}{h} \\ & f\left( x \right)=\frac{8xh+4{{h}^{2}}-5h}{h} \\ & \end{align} Solving further to get, \begin{align} & f\left( x \right)=\frac{h\left( 8x+4h-5 \right)}{h} \\ & =8x+h-5 \end{align} Thus, the simplified form of the provided expression is $8x+4h-5$.