91Ó°ÊÓ

Rearrange the inequality to 0 on one side as follows: \(\frac{x}{x+2} - 2 \geq 0\) which simplifies to \(\frac{x-2x-4}{x+2} \geq 0\) or \(\frac{-x-4}{x+2} \geq 0\).

The critical points will be found by setting the numerator and the denominator equal to zero. For \(-x-4 = 0\), x equals -4 and for \(x+2 = 0\), x equals -2. These are the critical points.

The intervals are (-∞, -4), (-4, -2), (-2, ∞). Choose a test point from each interval and substitute it into the simplified inequality expression. If the substitute results in a positive number (or zero), the inequality holds. If it results in a negative number, it does not hold. Test points could be -5, -3, and 0 respectively. At x=-5 (within (-∞, -4)), the value of the expression \(\frac{-x-4}{x+2}\) is positive, hence this interval is part of the solution set. At x=-3 (within (-4, -2)), the value of the expression is negative, hence this interval is not part of the solution. At x=0 (within (-2, ∞)), the value of the expression is negative, hence this interval is also not part of the solution. Note that the solution does not include the critical point -2, because at x=-2 the denominator equals 0, thus the expression is undefined.

The solution in interval notation is therefore (-∞, -4]. The graph on the number line will show a closed circle at -4 as the inequality includes equals to, and an open circle at -2 with the line extending towards -∞ as -2 doesn't belong to the solution.