Answer
The value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ is $-h-2x-2$.
Work Step by Step
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 $-h-2x-2$.
