Answer
The value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ with the functions $f\left( x \right)={{x}^{2}}-x-4$ and $g\left( x \right)=2x-6$ is $2x+h-1$.
Work Step by Step
Now, to find $f\left( x+h \right),$ substitute $x+h$ in place of x.
$x\to x+h$
$\begin{align}
& \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{\left\{ {{\left( x+h \right)}^{2}}-\left( x+h \right)-4 \right\}-\left( {{x}^{2}}-x-4 \right)}{h} \\
& =\frac{{{x}^{2}}+2xh+{{h}^{2}}-x-h-4-{{x}^{2}}+x+4}{h} \\
& =\frac{2xh+{{h}^{2}}-h}{h} \\
& =2x+h-1
\end{align}$
Therefore,
$\frac{f\left( x+h \right)-f\left( x \right)}{h}=2x+h-1$.
Hence, the value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ with the functions $f\left( x \right)={{x}^{2}}-x-4$ and $g\left( x \right)=2x-6$ is $2x+h-1$.
