To determine the intervals on which the function $f(x) = \frac{e^{-2x}}{-x}$ is decreasing, we need to analyze its derivative. Here is the step-by-step solution:

1. Rewrite the function: $f(x) = \frac{e^{-2x}}{-x}$

2. Find the derivative: We use the quotient rule for derivatives: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ Here, $u = e^{-2x}$ and $v = -x$. Compute $u'$ and $v'$:

   • $u' = \frac{d}{dx} e^{-2x} = -2e^{-2x}$
   • $v' = \frac{d}{dx} (-x) = -1$

   Substituting into the quotient rule: $f'(x) = \frac{(-2e^{-2x})(-x) - (e^{-2x})(-1)}{(-x)^2}$ Simplify the numerator: $f'(x) = \frac{2xe^{-2x} - e^{-2x}}{x^2}$ Factor $e^{-2x}$ from the numerator: $f'(x) = \frac{e^{-2x}(2x - 1)}{x^2}$

3. Determine when $f'(x)$ is negative: The function $f(x)$ is decreasing when $f'(x) < 0$. Analyze the numerator and denominator:

   • The denominator, $x^2$, is always positive for $x \neq 0$.
   • The numerator, $e^{-2x}(2x - 1)$, is determined by $2x - 1$, as $e^{-2x} > 0$ for all $x$.

   So, $f'(x) < 0$ when: $2x - 1 < 0 \quad \Rightarrow \quad x < \frac{1}{2}$

4. Consider the domain: The function $f(x)$ is undefined at $x = 0$. Hence, its domain is $x \in (-\infty, 0) \cup (0, \infty)$.

5. Conclusion: Combining the inequality and the domain:

   • $f(x)$ is decreasing on $(-\infty, 0) \cup (0, \frac{1}{2})$.

Let me know if you need further clarification or detailed steps!

Related Questions:

1. How do you determine if a function is increasing or decreasing using the derivative?
2. Can you explain the role of the quotient rule in derivative calculations?
3. What is the significance of the domain when analyzing $f(x)$?
4. Why is $e^{-2x}$ always positive, and how does that affect the derivative?
5. How does factoring help simplify derivative expressions?

Tip:

When analyzing whether a function is increasing or decreasing, always check both the numerator and denominator of the derivative and ensure the domain of the function is considered.