Write the domain of the function in interval notation.
-$$k ( x ) = \frac { 1 } { \sqrt { 2 x ^ { 2 } + 3 x - 54 } }$$
A) $( - \infty , - 6 ) \cup \left( \frac { 9 } { 2 } , \infty \right)$
B) $\left( - 6 , \frac { 9 } { 2 } \right)$
C) \{ \}
D) $( - \infty , \infty )$

Question

Quizplus · Accepted Answer

The domain of $k(x) = \frac{1}{\sqrt{2x^2 + 3x - 54}}$ is determined by the condition that the expression under the square root, $2x^2 + 3x - 54$, must be greater than 0 because the square root of a negative number is not real, and the denominator of a fraction cannot be 0. Factoring the quadratic expression gives $2x^2 + 3x - 54 = (2x - 9)(x + 6)$. Setting each factor greater than 0, we find the intervals where the expression is positive are $x < -6$ and $x > \frac{9}{2}$. Thus, the domain is $(- \infty , - 6 ) \cup \left( \frac { 9 } { 2 } , \infty \right)$.

Write the Domain of the Function in Interval Notation k(x)=12x2+3x−54k ( x ) = \frac { 1 } { \sqrt { 2 x ^ { 2 } + 3 x - 54 } }k(x)=2x2+3x−54​1​

Write the Domain of the Function in Interval Notation $k ( x ) = \frac { 1 } { \sqrt { 2 x ^ { 2 } + 3 x - 54 } }$