From what I can decipher, the problem involves analyzing a function $f(x) = -x - \frac{4}{x}$ and answering questions related to limits, derivatives, and possibly areas.

Due to the poor quality of the image, I can only attempt to answer parts of it.

1. Problem Description

From what I can decipher, the problem involves analyzing a function $f(x) = -x - \frac{4}{x}$ and answering questions related to limits, derivatives, and possibly areas.

2. Solution Steps

Let's analyze parts of the visible questions:

1. It appears to ask about the limit of $f(x)$ as $x$ approaches $0^+$ and $x$ approaches $+\infty$. It wants us to determine what happens to $f(x)$ in these cases.

As $x \to 0^+$, $f(x) = -x - \frac{4}{x} \approx -\frac{4}{x} \to -\infty$.

As $x \to +\infty$, $f(x) = -x - \frac{4}{x} \approx -x \to -\infty$.

2. It refers to the line $y = -x$ and seems to ask about asymptotes related to $(C)$.

Since $\lim_{x \to \infty} [f(x) - (-x)] = \lim_{x \to \infty} [-x - \frac{4}{x} + x] = \lim_{x \to \infty} -\frac{4}{x} = 0$, then $y=-x$ is an asymptote.

3. It asks us to find where $f'(x) < 0$. This means we need to find the derivative of $f(x)$:

$f(x) = -x - \frac{4}{x} = -x - 4x^{-1}$

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

To solve $f'(x) < 0$, we have $\frac{4 - x^2}{x^2} < 0$. Since $x > 0$, $x^2 > 0$, so we must have $4 - x^2 < 0$, meaning $x^2 > 4$. Thus $|x| > 2$, but since it says $x > 0$, we must have $x > 2$.

3. Final Answer

1. $\lim_{x \to 0^+} f(x) = -\infty$ and $\lim_{x \to +\infty} f(x) = -\infty$.

2. $y = -x$ is an asymptote of $f(x)$.

3. $f'(x) < 0$ when $x > 2$.