We are given the function $f(x) = x + 2 + \frac{4}{x-1}$ and asked to study its variations on the interval $[-4, 4]$. This means we need to find where the function is increasing, decreasing, or constant in this interval.

1. Problem Description

We are given the function $f(x) = x + 2 + \frac{4}{x-1}$ and asked to study its variations on the interval $[-4, 4]$. This means we need to find where the function is increasing, decreasing, or constant in this interval.

2. Solution Steps

First, find the derivative of the function $f(x)$:

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

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

Now, we want to find the critical points by setting $f'(x) = 0$:

$1 - \frac{4}{(x-1)^2} = 0$

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

$(x-1)^2 = 4$

$x-1 = \pm 2$

$x = 1 \pm 2$

So, $x = 3$ or $x = -1$.

The domain of $f(x)$ is all real numbers except $x=1$. Thus, $x=1$ is a vertical asymptote and also needs to be considered when studying the variations.

We now need to examine the sign of $f'(x)$ in the intervals $[-4, -1)$, $(-1, 1)$, $(1, 3)$, and $(3, 4]$.

Interval $[-4, -1)$: Let $x = -2$. Then $f'(-2) = 1 - \frac{4}{(-2-1)^2} = 1 - \frac{4}{9} = \frac{5}{9} > 0$. $f(x)$ is increasing on $[-4, -1)$.

Interval $(-1, 1)$: Let $x = 0$. Then $f'(0) = 1 - \frac{4}{(0-1)^2} = 1 - 4 = -3 < 0$. $f(x)$ is decreasing on $(-1, 1)$.

Interval $(1, 3)$: Let $x = 2$. Then $f'(2) = 1 - \frac{4}{(2-1)^2} = 1 - 4 = -3 < 0$. $f(x)$ is decreasing on $(1, 3)$.

Interval $(3, 4]$: Let $x = 4$. Then $f'(4) = 1 - \frac{4}{(4-1)^2} = 1 - \frac{4}{9} = \frac{5}{9} > 0$. $f(x)$ is increasing on $(3, 4]$.

In summary, $f(x)$ is increasing on $[-4, -1)$, decreasing on $(-1, 1)$, decreasing on $(1, 3)$, and increasing on $(3, 4]$. At $x=-1$, there is a local maximum. At $x=3$, there is a local minimum.

3. Final Answer

The function $f(x)$ is increasing on $[-4, -1)$, decreasing on $(-1, 1)$, decreasing on $(1, 3)$, and increasing on $(3, 4]$.

There is a local maximum at $x = -1$ and a local minimum at $x = 3$. There is a vertical asymptote at $x=1$.