The function you've provided is:

$f(x) = \frac{1}{\sqrt{1 - \sin(2x)}}$

This function can be analyzed in different ways, depending on what you're looking for, such as:

1. Domain: To determine where the function is defined, we need to ensure that the expression inside the square root, $1 - \sin(2x)$, is non-negative. Since $\sin(2x)$ can take values between -1 and 1, we must have: $1 - \sin(2x) \geq 0 \implies \sin(2x) \leq 1$ This inequality is always true for $\sin(2x)$. However, to avoid a zero denominator, $\sin(2x) \neq 1$, meaning that $2x \neq \frac{\pi}{2} + 2k\pi$ (where $k$ is an integer). Solving this gives: $x \neq \frac{\pi}{4} + k\pi \quad \text{(for integers } k)$

2. Range: The function takes values from $\frac{1}{\sqrt{1 - \sin(2x)}}$, and the minimum value of $\sin(2x)$ is -1, giving the range $\left( \frac{1}{\sqrt{1 + 1}}, \infty \right) = (1, \infty)$.

3. Derivative: The derivative of $f(x)$ can be found using the chain rule. Do you need help finding the derivative of this function?

Would you like to explore one of these topics or something else in particular?

5 Related Questions:

1. How can we find the critical points of $f(x)$?
2. What is the second derivative of $f(x)$?
3. How does the graph of $f(x)$ behave near discontinuities?
4. What are the asymptotes, if any, of this function?
5. How would you integrate this function over an interval?

Tip:

When dealing with trigonometric functions inside radicals, always ensure the expression inside the square root is non-negative to define the domain correctly.