Question

Find the domain of the real valued function of real variable: 

(iv)  \[f\left( x \right) = \frac{2x + 1}{x^2 - 9}\]

 

Answer 1

(iv) Given: 

\[f\left( x \right) = \frac{2x + 1}{x^2 - 9}\]

Domain of f :
Clearly,  f (x) is defined for all  x ∈ R except for x² -9 ≠  0, i.e. x = ± 3.
At x = -3, 3,  f (x) takes the intermediate form

\[\frac{1}{0} .\]  Hence, domain ( f ) = R -{ - 3, 3 }.