How do you find $g\left( 3 \right)$ given that $g\left( x \right)=-2{{x}^{2}}-9$?

Hint: We try to explain the function and its value at a particular point. For the given function $g\left( x \right)=-2{{x}^{2}}-9$, we need to find the value of the function at $x=3$. This means we are finding $g\left( 3 \right)$ for that particular point.

Complete step-by-step solution:
The given problem is about a given function $g\left( x \right)=-2{{x}^{2}}-9$ and the value of the function at that point of $x=3$ to find $g\left( 3 \right)$.
So, in those cases we find the value of the function by putting the point $x=3$ in the function of $g\left( x \right)=-2{{x}^{2}}-9$.
$g\left( 3 \right)=-2\times {{3}^{2}}-9$.
Now we apply the common binary operations to find the value for $g\left( 3 \right)$.
We need to find the square value of 3 which gives ${{3}^{2}}=3\times 3=9$.
We then multiply the value of the square with 2 and get $9\times 2=18$.
Now the equation becomes $g\left( 3 \right)=-18-9$.
We have two variables $-18$ and $-9$. The binary operation between them is addition.
This gives $g\left( 3 \right)=-18-9=-27$.
Therefore, for the given function $g\left( x \right)=-2{{x}^{2}}-9$, the value of $g\left( 3 \right)$ is $-27$.

Note: The domain of the function has to consist of the value of $x=3$. The function can give values for only its domain and the values of the function $g\left( x \right)=-2{{x}^{2}}-9$ gives the range for that particular function. In this case $-27$ is an element of the range whereas 3 was an element of the domain.