Answer
The value of $\frac{f\left( 2+h \right)-f\left( 2 \right)}{h}$ is $ h+5$.
Work Step by Step
Consider the function $ f\left( x \right)={{x}^{2}}+x $,
Find the value of $\frac{f\left( 2+h \right)-f\left( 2 \right)}{h}$ using the definition of the function.
$\begin{align}
& \frac{f\left( 2+h \right)-f\left( 2 \right)}{h}=\frac{{{\left( 2+h \right)}^{2}}+\left( 2+h \right)-\left( {{2}^{2}}+2 \right)}{h} \\
& =\frac{4+{{h}^{2}}+4h+2+h-\left( 4+2 \right)}{h} \\
& =\frac{{{h}^{2}}+5h+6-6}{h} \\
& =\frac{{{h}^{2}}+5h}{h}
\end{align}$
Factoring out h from the numerator,
$\begin{align}
& \frac{f\left( 2+h \right)-f\left( 2 \right)}{h}=\frac{h\left( h+5 \right)}{h} \\
& =h+5
\end{align}$
Thus, the value of $\frac{f\left( 2+h \right)-f\left( 2 \right)}{h}$ is $ h+5$.
