We are given two functions: $f(x) = x + 2 + \frac{4}{x-1}$ and $g(x) = x \ln x$. We need to study the variations of $f(x)$ on the interval $[-4, 4]$ and the variations of $g(x)$ on the interval $[2, 14]$. This means determining where the functions are increasing, decreasing, and finding any critical points.

1. Problem Description

We are given two functions: $f(x) = x + 2 + \frac{4}{x-1}$ and $g(x) = x \ln x$. We need to study the variations of $f(x)$ on the interval $[-4, 4]$ and the variations of $g(x)$ on the interval $[2, 14]$. This means determining where the functions are increasing, decreasing, and finding any critical points.

2. Solution Steps

Part 1: Study the variations of $f(x) = x + 2 + \frac{4}{x-1}$ on $[-4, 4]$.

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

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

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

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

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

$(x-1)^2 = 4$

$x-1 = \pm 2$

$x = 1 \pm 2$

$x = 3$ or $x = -1$

The critical points are $x = 3$ and $x = -1$. Note that $x=1$ is not in the domain of $f(x)$.

Now, analyze the sign of $f'(x)$ on the intervals $[-4, -1)$, $(-1, 1)$, $(1, 3)$, and $(3, 4]$.

- For $x \in [-4, -1)$, let's take $x = -2$. Then $f'(-2) = 1 - \frac{4}{(-2-1)^2} = 1 - \frac{4}{9} = \frac{5}{9} > 0$. So $f(x)$ is increasing on $[-4, -1)$.

- For $x \in (-1, 1)$, let's take $x = 0$. Then $f'(0) = 1 - \frac{4}{(0-1)^2} = 1 - 4 = -3 < 0$. So $f(x)$ is decreasing on $(-1, 1)$.

- For $x \in (1, 3)$, let's take $x = 2$. Then $f'(2) = 1 - \frac{4}{(2-1)^2} = 1 - 4 = -3 < 0$. So $f(x)$ is decreasing on $(1, 3)$.

- For $x \in (3, 4]$, let's take $x = 3.5$. Then $f'(3.5) = 1 - \frac{4}{(3.5-1)^2} = 1 - \frac{4}{6.25} = 1 - \frac{16}{25} = \frac{9}{25} > 0$. So $f(x)$ is increasing on $(3, 4]$.

In summary:

- $f(x)$ is increasing on $[-4, -1)$

- $f(x)$ is decreasing on $(-1, 1)$

- $f(x)$ is decreasing on $(1, 3)$

- $f(x)$ is increasing on $(3, 4]$

Part 2: Study the variations of $g(x) = x \ln x$ on $[2, 14]$.

First, find the derivative of $g(x)$:

$g'(x) = \frac{d}{dx} (x \ln x) = \ln x + x \cdot \frac{1}{x} = \ln x + 1$

Now, find the critical points by setting $g'(x) = 0$:

$\ln x + 1 = 0$

$\ln x = -1$

$x = e^{-1} = \frac{1}{e} \approx 0.368$

Since $\frac{1}{e}$ is not in the interval $[2, 14]$, we don't need to consider it.

Now, analyze the sign of $g'(x)$ on the interval $[2, 14]$.

Since $x \ge 2$, $\ln x > 0$, so $\ln x + 1 > 1 > 0$.

Thus, $g'(x) > 0$ for all $x \in [2, 14]$.

Therefore, $g(x)$ is increasing on $[2, 14]$.

3. Final Answer

For $f(x) = x + 2 + \frac{4}{x-1}$ on $[-4, 4]$:

- $f(x)$ is increasing on $[-4, -1)$.

- $f(x)$ is decreasing on $(-1, 1)$.

- $f(x)$ is decreasing on $(1, 3)$.

- $f(x)$ is increasing on $(3, 4]$.

For $g(x) = x \ln x$ on $[2, 14]$:

- $g(x)$ is increasing on $[2, 14]$.