Answer
The value of $f\left( -x \right)$ is $\underline{{{x}^{2}}+x-4}$ and the function is neither odd nor even.
Work Step by Step
We know that the function is said to be even when it satisfies $f\left( x \right)=f\left( -x \right)$ and the function is said to be odd when it satisfies $f\left( x \right)=-f\left( -x \right)$.
If the function satisfies neither of the two conditions, then the function is neither even nor odd.
To calculate $f\left( -x \right)$ substitute x with $-x$ in the function as follows:
$\begin{align}
& f\left( -x \right)={{\left( -x \right)}^{2}}-\left( -x \right)-4 \\
& ={{x}^{2}}+x-4
\end{align}$
It is clear that neither $f\left( -x \right)\ne f\left( x \right)$ nor $f\left( -x \right)=-f\left( x \right)$.
Therefore, $f$ is neither an even nor an odd function.
