What is the domain of the function �

The function in question is:

$y = \frac{2x}{\sqrt{5x^2 - 4}}$

To find the domain, we need to consider the following conditions:

1. Square Root Condition:

The expression inside the square root must be non-negative, because the square root of a negative number is not a real number. Therefore, we need: $5x^2 - 4 \geq 0$

2. Solve the Inequality:

$5x^2 \geq 4$ Divide both sides by 5: $x^2 \geq \frac{4}{5}$ Taking the square root on both sides: $|x| \geq \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$ This implies: $x \leq -\frac{2\sqrt{5}}{5} \quad \text{or} \quad x \geq \frac{2\sqrt{5}}{5}$

3. Denominator Condition:

The denominator cannot be zero because division by zero is undefined. Therefore, we must have: $5x^2 - 4 \neq 0$ Which simplifies to: $x^2 \neq \frac{4}{5}$ This does not change our interval but confirms the boundary values are excluded.

Conclusion:

The domain of the function $y = \frac{2x}{\sqrt{5x^2 - 4}}$ is:

$(-\infty, -\frac{2\sqrt{5}}{5}) \cup (\frac{2\sqrt{5}}{5}, \infty)$

This excludes the interval $-\frac{2\sqrt{5}}{5} \leq x \leq \frac{2\sqrt{5}}{5}$, where the expression inside the square root would be non-positive.

Would you like more details, or do you have any other questions?

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Here are 5 related questions:

1. What is the domain of the function $y = \frac{1}{\sqrt{9x^2 - 16}}$?
2. How does the domain change if the function is $y = \frac{3x + 1}{\sqrt{7 - x^2}}$?
3. What happens to the domain if the square root is in the numerator instead?
4. How would you determine the domain if the expression inside the square root is a cubic polynomial?
5. Can you determine the domain of a function with a square root in both the numerator and denominator?

Tip: When solving inequalities involving square roots, it's important to remember that the expression under the square root must be greater than zero when in the denominator.