Find the domain of $f ( x ) = \frac { x ^ { 2 } - 9 } { \sqrt { x + 3 } }$
A) Domain: $[ - 3 , \infty )$
B) Domain: $( - 3 , \infty )$
C) Domain: $( - \infty , \infty )$
D) Domain: $( - \infty , - 3 ) \cup ( - 3 , \infty )$

Question

Quizplus · Accepted Answer

The domain of $f(x) = \frac{x^2 - 9}{\sqrt{x + 3}}$ is determined by the conditions under which the expression is defined. The denominator, $\sqrt{x + 3}$, requires $x + 3 > 0$ (since the square root of a negative number is not real), which simplifies to $x > -3$. Therefore, the domain is all real numbers greater than -3, which is represented by the interval $(-3, \infty)$.

Find the Domain Of f(x)=x2−9x+3f ( x ) = \frac { x ^ { 2 } - 9 } { \sqrt { x + 3 } }f(x)=x+3​x2−9​

Find the Domain Of $f ( x ) = \frac { x ^ { 2 } - 9 } { \sqrt { x + 3 } }$