Answer with explanation: To find the values of x where the f(x) is discontinuous, we have to set the denominator equal to zero, doing this gives:
[tex]\begin{gathered} f(x)=\frac{x^2+10+9}{x^2-81}\Rightarrow x^2-81=0 \\ x=\sqrt[]{81}=9 \\ x=9 \end{gathered}[/tex]
The f(x) is discontinuous at x = 9, following graph confirms it:
In conclusion, discontinuity is non-removable.
