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Finding the critical points is the first and foremost step in solving an inequality involving a rational function. They are found where either the numerator or the denominator is zero. These points are crucial because they indicate potential changes in the inequality sign.

To identify these points, set each part of the fraction separately to zero:

• For the numerator \((x-1)^2\), it is zero when \(x = 1\). However, it's important to notice that since this is a square, it's always non-negative, and it only equals zero at \(x = 1\).
• For the denominator \((x+1)(x+2)\), set it equal to zero. This yields \(x = -1\) and \(x = -2\) as critical points.

These critical points are key to dividing the number line into regions or intervals where we will further test the inequality.

The numerator and denominator in a rational inequality serve different roles in solving the inequality.

The **numerator**: In this problem, the numerator \((x-1)^2\) is of particular interest because it provides a zero at \(x = 1\). With squares, we don't worry about negative outcomes—it’s either zero or positive, which impacts how we perceive the inequality at that point.
The **denominator**: The denominator \((x+1)(x+2)\) is equally crucial because it provides zeros at \(x = -1\) and \(x = -2\). At these points, the function is undefined (division by zero), and we need to exclude them from the solution set. Additionally, the sign depends on the expression being positive or negative in a given interval. Simply put, each zero in the denominator flips the sign of the inequality in adjacent intervals, hence influencing the solution set.

After determining the critical points, we proceed with interval testing. This process helps us know where the original inequality holds true. The critical points divide the coordinate axis into several intervals.

We choose any convenient "test point" from each interval. Plug this point into the original inequality to check whether the fraction's value is positive or negative.

• Interval \((-\infinity, -2)\) with test point \(x = -3\): yields a positive fraction.
• Interval \((-2, -1)\) with test point \(x = -1.5\): results in a negative fraction, meaning the inequality does not hold here.
• Interval \((-1, 1)\) with test point \(x = 0\): gives a positive result again.
• Interval \((1, \infinity)\) with test point \(x = 2\): also results in a positive fraction.

This method of interval testing ensures that we accurately find where the inequality is satisfied.

Once all intervals have been tested, the solution set is determined by compiling intervals where the inequality holds true.

From the earlier tests, the inequality \(\frac{(x-1)^2}{(x+1)(x+2)} > 0\) is true in the intervals \((-\infinity, -2)\), \((-1, 1)\), and \((1, \infinity)\).
We must remember to exclude points where the denominator becomes zero, which are \(x = -1\) and \(x = -2\). Excluding these ensures that we only include sections where the function is well-defined.
The final solution set is created using union notation: \((-\infinity, -2) \cup (-1, 1) \cup (1, \infinity)\). This represents all the \(x\)-values that satisfy the original inequality.