## Precalculus (6th Edition) Blitzer

The value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ is $-h-2x-2$.
We have to obtain the value of $f\left( x+h \right)$; replace $x$ in the provided function $f\left( x \right)$ by $\left( x+h \right)$. Since, $f\left( x \right)=-{{x}^{2}}-2x+1$ then \begin{align} & f\left( x+h \right)=-{{\left( x+h \right)}^{2}}-2\left( x+h \right)+1 \\ & =-{{x}^{2}}-{{h}^{2}}-2xh-2x-2h+1 \end{align} Then, for $f\left( x+h \right)-f\left( x \right)$: \begin{align} & f\left( x+h \right)-f\left( x \right)=\left( -{{x}^{2}}-{{h}^{2}}-2xh-2x-2h+1 \right)-\left( -{{x}^{2}}-2x+1 \right) \\ & =-{{x}^{2}}-{{h}^{2}}-2xh-2x-2h+1+{{x}^{2}}+2x-1 \\ & =-h\left( h+2x+2 \right) \end{align} Next, divide $f\left( x+h \right)-f\left( x \right)$ by ‘h’ to get $\frac{f\left( x+h \right)-f\left( x \right)}{h}$: \begin{align} & \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{-h\left( h+2x+2 \right)}{h} \\ & =-h-2x-2 \end{align} Thus, the value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ is $-h-2x-2$.