The given function $f(x)$ is expressed as:

$f(x) = \frac{e^{-x}}{e^x + 2e^{-x} - 3}$

You are asking for the domain of the function $f(x)$, denoted as $Df$, which refers to the set of all $x$ values for which the function is defined.

To find the domain:

The function is a rational function, and for rational functions, the domain consists of all real numbers except those that make the denominator zero.

We need to find when the denominator becomes zero:

$e^x + 2e^{-x} - 3 = 0$

Multiplying the entire equation by $e^x$ (to eliminate the negative exponent), we get:

$e^{2x} + 2 - 3e^x = 0$

This is a quadratic equation in terms of $e^x$. Let $u = e^x$, then the equation becomes:

$u^2 - 3u + 2 = 0$

Factoring the quadratic equation:

$(u - 1)(u - 2) = 0$

So, $u = 1$ or $u = 2$. Since $u = e^x$, we set $e^x = 1$ and $e^x = 2$, which gives:

1. $e^x = 1$ implies $x = 0$
2. $e^x = 2$ implies $x = \ln(2)$

Thus, the function is undefined at $x = 0$ and $x = \ln(2)$. Therefore, the domain of $f(x)$ is:

$Df = \mathbb{R} \setminus \{0, \ln(2)\}$

Let me know if you'd like further explanation!

Here are 5 related questions:

1. How do we handle exponents and logarithms in solving for the domain?
2. How do you determine the domain of a rational function generally?
3. What is the significance of removing values from the domain?
4. How does multiplying by $e^x$ simplify the equation?
5. How would you graph the function to visualize its domain?

Tip: When solving for the domain of rational functions, always check for values that make the denominator zero, as those are the points where the function becomes undefined.