We want to find the value of x that is not in the domain of the function;
[tex]h(x)=\frac{x^2+5x-14}{x^2-81}[/tex]
The value of x that will make h undefined is the value that makes the denominator zero.
Therefore;
[tex]\begin{gathered} x^2-81=0 \\ x^2=81 \\ x=\pm9_{} \end{gathered}[/tex]
Therefore, the values not in the domain is +9 and -9
