Originally Posted by
topsquark I'll show you how to do the first. The second is done the same way.
$\displaystyle x(x - 1)(x + 2) > 0$
First we want to get everything onto one side of the equation. That's already done here.
Next we want to identify "critical points." This is where the LHS of the equation is 0, or where the denominator is 0. Here we have no denominator, so our critical points are x = -2, 0, 1.
Now we want to break the real line into pieces, given by the critical points. In this case we have 4 intervals:
$\displaystyle ( - \infty, -2), (-2, 0), (0, 1), (1, \infty)$
Now we want to test values of x from each interval in the inequality to see which ones give the correct result:
$\displaystyle ( - \infty, -2)$ gives $\displaystyle x(x - 1)(x + 2) < 0$ (No!)
$\displaystyle (-2, 0)$ gives $\displaystyle x(x - 1)(x + 2) > 0$ (Yes!)
$\displaystyle (0, 1)$ gives $\displaystyle x(x - 1)(x + 2) < 0$ (No!)
$\displaystyle (1, \infty )$ gives $\displaystyle x(x - 1)(x + 2) > 0$ (Yes!)
Only two of these intervals have x values that work in the inequality. These intervals are the solution set. Thus the answer is $\displaystyle ( -2, 0) \cup (1, \infty )$.
The answer to the second problem is $\displaystyle (-1, 4] $. (The 4 is included in the solution set because we have a $\displaystyle \leq$ symbol. Why isn't -1 included in the solution set?)
-Dan