To determine the domain of the function
[tex]\[
f(x) = \frac{1}{\sqrt{5x^2 + 9x - 2}},
\][/tex]
we need to identify the values of [tex]\( x \)[/tex] for which the function is defined. Specifically, the expression under the square root in the denominator must be positive, because the square root of a negative number is not real, and division by zero is undefined. Therefore, we need to solve the inequality:
[tex]\[
5x^2 + 9x - 2 > 0.
\][/tex]
To solve this, we will first find the roots of the corresponding equation:
[tex]\[
5x^2 + 9x - 2 = 0.
\][/tex]
We use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\][/tex]
where [tex]\( a = 5 \)[/tex], [tex]\( b = 9 \)[/tex], and [tex]\( c = -2 \)[/tex]. Plugging in these values, we get:
[tex]\[
x = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 5 \cdot (-2)}}{2 \cdot 5}.
\][/tex]
Calculate the discriminant:
[tex]\[
9^2 - 4 \cdot 5 \cdot (-2) = 81 + 40 = 121.
\][/tex]
So, we have:
[tex]\[
x = \frac{-9 \pm \sqrt{121}}{10} = \frac{-9 \pm 11}{10}.
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{-9 + 11}{10} = \frac{2}{10} = \frac{1}{5},
\][/tex]
[tex]\[
x = \frac{-9 - 11}{10} = \frac{-20}{10} = -2.
\][/tex]
These roots, [tex]\( x = \frac{1}{5} \)[/tex] and [tex]\( x = -2 \)[/tex], divide the real number line into three intervals: [tex]\( (-\infty, -2) \)[/tex], [tex]\( (-2, \frac{1}{5}) \)[/tex], and [tex]\( (\frac{1}{5}, \infty) \)[/tex].
Next, we need to determine the sign of [tex]\( 5x^2 + 9x - 2 \)[/tex] in each of these intervals. We can use test points from each interval:
1. For [tex]\( x \in (-\infty, -2) \)[/tex], choose [tex]\( x = -3 \)[/tex]:
[tex]\[
5(-3)^2 + 9(-3) - 2 = 45 - 27 - 2 = 16 \quad (\text{positive}).
\][/tex]
2. For [tex]\( x \in (-2, \frac{1}{5}) \)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[
5(0)^2 + 9(0) - 2 = -2 \quad (\text{negative}).
\][/tex]
3. For [tex]\( x \in (\frac{1}{5}, \infty) \)[/tex], choose [tex]\( x = 1 \)[/tex]:
[tex]\[
5(1)^2 + 9(1) - 2 = 5 + 9 - 2 = 12 \quad (\text{positive}).
\][/tex]
From this analysis, we see that the quadratic expression [tex]\( 5x^2 + 9x - 2 \)[/tex] is positive in the intervals [tex]\( (-\infty, -2) \)[/tex] and [tex]\( (\frac{1}{5}, \infty) \)[/tex]. Hence, the function [tex]\( f(x) \)[/tex] is defined, and therefore its domain is:
[tex]\[
(-\infty, -2) \cup \left(\frac{1}{5}, \infty\right).
\][/tex]
So, the domain of the function is
[tex]\[
\boxed{(-\infty, -2) \cup \left(\frac{1}{5}, \infty\right)}.
\][/tex]
