Let's analyze the function provided:

$f(x) = \frac{5}{1 + 4e^{-2x}}$

and address the given questions:

(a) Find the horizontal asymptotes

Horizontal asymptotes are determined by examining the behavior of $f(x)$ as $x \to \infty$ and $x \to -\infty$.

1. As $x \to \infty$:

   • $e^{-2x} \to 0$, so the denominator approaches $1 + 4(0) = 1$.
   • Thus, $f(x) \to \frac{5}{1} = 5$.

2. As $x \to -\infty$:

   • $e^{-2x} \to \infty$, so the denominator approaches $1 + 4(\infty) = \infty$.
   • Thus, $f(x) \to \frac{5}{\infty} = 0$.

Horizontal Asymptotes:

• $y = 5$ as $x \to \infty$
• $y = 0$ as $x \to -\infty$

(b) Find the domain and range

1. Domain: The denominator $1 + 4e^{-2x}$ is always positive because $e^{-2x} > 0$ for all real $x$. There are no restrictions (e.g., division by zero or undefined expressions).

   • Domain: $(-\infty, \infty)$.

2. Range:

   • As $x \to \infty$, $f(x) \to 5$.
   • As $x \to -\infty$, $f(x) \to 0$.
   • Since $f(x)$ is always positive and smoothly transitions between these values, the range is:
     • Range: $(0, 5)$.

(c) Describe the end behavior

1. As $x \to \infty$, $f(x) \to 5$. The function levels off at the horizontal asymptote $y = 5$.
2. As $x \to -\infty$, $f(x) \to 0$. The function approaches the horizontal asymptote $y = 0$ but never touches it.

End behavior:

• $f(x) \to 5$ as $x \to \infty$
• $f(x) \to 0$ as $x \to -\infty$

Let me know if you need detailed explanations or additional assistance!

Related Questions:

1. What is the significance of the exponential decay term $e^{-2x}$ in this function?
2. How does $f(x)$ behave near $x = 0$?
3. Can you graph the function $f(x)$ to visually verify the asymptotes and range?
4. What transformations are applied to the base exponential function $e^{-x}$ to form this function?
5. How does this function model real-world phenomena, such as logistic growth?

Tip: When analyzing asymptotes, always focus on the dominant terms of the function as $x \to \pm \infty$.