Find the Domain of \(f(x)=\frac{2x+1}{x^2-9}\)

Solution

Given: $$ f(x)=\frac{2x+1}{x^2-9} $$

The denominator cannot be zero.

$$ x^2-9\ne0 $$

$$ (x-3)(x+3)\ne0 $$

Therefore, $$ x\ne3 \quad \text{and} \quad x\ne-3 $$

Hence, all real numbers except \(-3\) and \(3\) are allowed.

Therefore, the domain is: $$ \mathbb{R}-\{-3,3\} $$

or $$ (-\infty,-3)\cup(-3,3)\cup(3,\infty) $$