Question

Solve the inequality and write the solution set in interval notation. \((x-1)^{2}(x+2)^{3} \geq 0\)

Short answer

The solution to the inequality \((x-1)^{2}(x+2)^{3} \geq 0\) is all real numbers i.e., \(-\infty < x < \infty\) in interval notation.

Step-by-step solution

Determine the critical points

First, determine the solutions to the equation \((x-1)^{2}(x+2)^{3} = 0\). To find these, set both \((x-1)^{2}\) and \((x+2)^{3}\) equal to zero separately, as the entire expression being zero can only be true if one of these factors is zero. This gives us the solutions \(x = 1\) and \(x=-2\), which are the critical points.

Partition the x-axis and test the regions

Now, consider the regions partitioned by these critical points on the x-axis: \(-\infty\) to \(-2\), \(-2\) to \(1\), and \(1\) to \(\infty\). We choose test points within these regions, for example, \(-3\) in [-\(\infty\), -2], \(-1\) in (-2, 1) and 2 in (1, \(\infty\)), and check whether ∃x in these intervals which satisfies the inequality \((x-1)^{2}(x+2)^{3} \geq 0\).

Write the solution in interval notation

As both factors are squared or cubed, they will always be positive. Therefore, the entire expression is non-negative i.e., \((x-1)^{2}(x+2)^{3} \geq 0\) holds for all real numbers. In terms of the interval notation, the solution set of \((x-1)^{2}(x+2)^{3} \geq 0\) is \(-\infty < x < \infty\).