The given function is:

$f(x) = \frac{1}{(x+1)^2}, \quad \text{where} \quad x \geq 0$

This is a rational function that involves a denominator of $(x+1)^2$. It is defined for $x \geq 0$ as specified.

Key features of this function:

1. Domain: The function is defined for $x \geq 0$. The expression in the denominator, $(x+1)^2$, is always positive and does not cause division by zero for $x \geq 0$.

2. Asymptotes:

   • Horizontal asymptote: As $x \to \infty$, the function approaches 0 because the denominator grows much faster than the numerator. Thus, $y = 0$ is a horizontal asymptote.
   • No vertical asymptote: Since the denominator does not become zero for $x \geq 0$, there is no vertical asymptote.

3. Behavior near $x = 0$: When $x = 0$, $f(0) = \frac{1}{(0+1)^2} = 1$. The function is decreasing as $x$ increases from 0 onwards.

4. Monotonicity: The function is strictly decreasing for all $x \geq 0$ because as $x$ increases, the denominator increases, making $f(x)$ smaller.

Would you like a detailed derivative analysis or more information on another feature of the function?

Relative Questions:

1. What is the derivative of $f(x) = \frac{1}{(x+1)^2}$?
2. How would you find the minimum value of the function $f(x)$?
3. Can you explain the concavity of this function based on the second derivative?
4. What is the behavior of the function as $x \to 0^+$?
5. How would you integrate the function $f(x)$ over a given interval?

Tip:

Always check for asymptotic behavior in rational functions—both horizontal and vertical asymptotes give important information about the limits of the function.