The problem asks us to use the given graph to determine which of the following intervals is part of the solution set of the inequality $2x + 1 > \frac{x+3}{x-1}$.

1. Problem Description

The problem asks us to use the given graph to determine which of the following intervals is part of the solution set of the inequality $2x + 1 > \frac{x+3}{x-1}$.

2. Solution Steps

First, let's rewrite the inequality:

$2x + 1 > \frac{x+3}{x-1}$

$2x + 1 - \frac{x+3}{x-1} > 0$

$\frac{(2x+1)(x-1) - (x+3)}{x-1} > 0$

$\frac{2x^2 - 2x + x - 1 - x - 3}{x-1} > 0$

$\frac{2x^2 - 2x - 4}{x-1} > 0$

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

$\frac{2(x-2)(x+1)}{x-1} > 0$

$\frac{(x-2)(x+1)}{x-1} > 0$

Let $f(x) = \frac{(x-2)(x+1)}{x-1}$. We need to find the intervals where $f(x) > 0$.

The critical points are $x = -1, 1, 2$.

We will test the intervals $(-\infty, -1), (-1, 1), (1, 2), (2, \infty)$.

Interval $(-\infty, -1)$, test $x = -2$:

$f(-2) = \frac{(-2-2)(-2+1)}{-2-1} = \frac{(-4)(-1)}{-3} = \frac{4}{-3} < 0$

Interval $(-1, 1)$, test $x = 0$:

$f(0) = \frac{(0-2)(0+1)}{0-1} = \frac{(-2)(1)}{-1} = 2 > 0$

Interval $(1, 2)$, test $x = 1.5$:

$f(1.5) = \frac{(1.5-2)(1.5+1)}{1.5-1} = \frac{(-0.5)(2.5)}{0.5} = -2.5 < 0$

Interval $(2, \infty)$, test $x = 3$:

$f(3) = \frac{(3-2)(3+1)}{3-1} = \frac{(1)(4)}{2} = 2 > 0$

Therefore, the solution set is $(-1, 1) \cup (2, \infty)$.

Looking at the graph, $2x+1$ is a straight line and $\frac{x+3}{x-1}$ is a rational function with a vertical asymptote at $x=1$.

The inequality $2x + 1 > \frac{x+3}{x-1}$ means we are looking for the regions where the line is above the rational function.

The straight line intersects the rational function at $x=-1$ and $x=2$.

From the graph, we can see that $2x+1 > \frac{x+3}{x-1}$ in the interval $(-1, 1)$ and in the interval $(2, \infty)$.

Among the options:

$x \ge 2$ means $[2, \infty)$, which is included in the solution.

$-1 < x < 2$ means $(-1, 2)$, which is not entirely included in the solution. Only $(-1, 1)$ is part of the solution.

$x > 2$ means $(2, \infty)$, which is part of the solution.

$x < -1$ means $(-\infty, -1)$, which is not part of the solution.

Since we want to determine which of the following intervals *is part of* the solution set, both $x \ge 2$ and $x>2$ are correct. However, it seems like the answer is most likely to be $x>2$.

3. Final Answer

x > 2