Answer
The required solution is $\frac{f\left( x+h \right)-f\left( x \right)}{h}=2h+4x-1$.
Work Step by Step
We have the functions $f\left( x \right)=2{{x}^{2}}-x-1$ and $ g\left( x \right)=1-x $:
Therefore, compute $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ as given below:
$\begin{align}
& \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{2{{\left( x+h \right)}^{2}}-\left( x+h \right)-1-\left( 2{{x}^{2}}-x-1 \right)}{h} \\
& =\frac{2{{\left( x+h \right)}^{2}}-\left( x+h \right)-1-\left( 2{{x}^{2}}-x-1 \right)}{h} \\
& =\frac{2{{h}^{2}}+4xh-h}{h} \\
& =2h+4x-1
\end{align}$
Thus, $\frac{f\left( x+h \right)-f\left( x \right)}{h}=2h+4x-1$.
