Question

Solve each inequality algebraically. $$\frac{x\left(x^{2}+1\right)(x-2)}{(x-1)(x+1)} \geq 0$$

Short answer

\((-\infty, -1)\cup[0, 1)\cup [2, +\infty)\)

Step-by-step solution

Identify the critical points

The critical points are values of x that make the numerator or denominator zero. Set each factor in the numerator and the denominator equal to zero and solve for x. 1. \(x = 0\) from \(x\) in the numerator.2. \(x = -i \text{ and } x = i\) from \(x^2 + 1\) in the numerator (these are imaginary and won't affect the real solution domain).3. \(x = 2\) from \(x - 2\) in the numerator.4. \(x = 1\) from \(x - 1\) in the denominator.5. \(x = -1\) from \(x + 1\) in the denominator. So, the critical points are \(x = 0, 1, -1, \text{ and } 2\).

Determine sign changes across the intervals

Create a number line and plot the critical points \(x = -1, 0, 1\), and \(2\). They divide the number line into the intervals: \((-\infty, -1)\), \((-1, 0)\), \((0, 1)\), \((1, 2)\), and \((2, +\infty)\). Test a point from each interval in the inequality to determine if it is positive or negative.

Test points in each interval

Choose test points: -2, -0.5, 0.5, 1.5, and 3. Plug these points into the expression \(\frac{x(x^2+1)(x-2)}{(x-1)(x+1)}\) and determine the sign of the result.1. For \(x = -2\): \( \frac{-2((-2)^2+1)(-2-2)}{(-2-1)(-2+1)} = \frac{-2(5)(-4)}{-3(-1)} = \frac{40}{3} > 0\), so the expression is positive on \((-\infty, -1)\).2. For \(x = -0.5\): \( \frac{-0.5((-0.5)^2+1)(-0.5-2)}{(-0.5-1)(-0.5+1)} = \frac{-0.5(1.25)(-2.5)}{(-1.5)(0.5)} = \frac{1.5625}{-0.75} < 0\), so it is negative on \((-1, 0)\).3. For \(x = 0.5\): \( \frac{0.5((0.5)^2 + 1)(0.5-2)}{ (0.5-1)(0.5+1)} = \frac{0.5(1.25)(-1.5)}{-0.5(1.5)} = \frac{-0.9375}{-0.75} > 0\), so it is positive on \((0, 1)\).4. For \(x = 1.5\): \( \frac{1.5((1.5)^2 + 1)(1.5-2)}{ (1.5-1)(1.5+1)} = \frac{1.5(3.25)(-0.5)}{(0.5)(2.5)} = \frac{-2.4375}{1.25} < 0\), so it is negative on \((1, 2)\).5. For \(x = 3\): \( \frac{3((3)^2+1)(3-2)}{(3-1)(3+1)} = \frac{3(10)(1)}{(2)(4)} = \frac{30}{8} > 0\), so it is positive on \((2, +\infty)\).

Determine which points to include

Inequality is \(\geq 0\), so include the points where the expression is zero but denominator cannot be zero. This occurs at \(x = 0\) and \(x = 2\) but not \(x = -1\) and \(x = 1\) as they make the denominator zero.

Combine intervals

Combining all the intervals where the inequality holds true and including the points where the expression is zero but excluding where denominator zero.

Key concepts

critical points

In any polynomial inequality, identifying the critical points is the first essential step. Critical points are the values that make either the numerator or denominator equal to zero. These values divide the number line into distinct intervals where the polynomial's sign might change. For example, in the given inequality,\[\frac{x\big(x^{2}+1\big)(x-2)}{(x-1)(x+1)} \geq 0\], we find the critical points by solving for x in each factor:
- \(\boldsymbol{x = 0}\) from \(\boldsymbol{x}\) in the numerator.
- \(\boldsymbol{x = 1}\) from \(\boldsymbol{x - 1}\) in the denominator.
- \(\boldsymbol{x = -1}\) from \(\boldsymbol{x + 1}\) in the denominator.
- \(\boldsymbol{x = 2}\) from \(\boldsymbol{x - 2}\) in the numerator.
Imaginary roots from \(\boldsymbol{x^2 + 1 = 0}\) are not considered since they are not real.
Once identified, these critical points are marked on a number line to determine where to test the intervals created by these points.

interval testing

After marking the critical points on the number line, the next step is interval testing. This involves selecting a number within each interval and substituting it back into the inequality to determine the sign of the expression in that interval. The intervals for our inequality divided by the critical points are:
- \((-\infty, -1)\)
- \((-1, 0)\)
- \((0, 1)\)
- \((1, 2)\)
- \((2, +\infty)\)
Test a number from each interval in the original expression \[\frac{x(x^2+1)(x-2)}{(x-1)(x+1)}\]. Here's how:
- For \(x = -2\), the expression is positive.
- For \(x = -0.5\), the expression is negative.
- For \(x = 0.5\), the expression is positive.
- For \(x = 1.5\), the expression is negative.
- For \(x = 3\), the expression is positive.
By testing these points, we can determine whether the polynomial on each interval is positive or negative. This helps us understand where the original inequality holds true (in this case, where the expression is greater than or equal to zero).

algebraic inequalities

In solving polynomial inequalities, the overall goal is to identify the intervals where the inequality satisfies the given condition. Algebraic inequalities like our example require a systematic approach to determine where the expression is positive or non-negative. Once you test the intervals, combine those that meet the inequality's requirement and include endpoints if specified to do so (like in \(\geq 0\)).
Finally, we summarize the solution set by recognizing the correct intervals.
From our testing:

• The expression is positive on \(( -\infty, -1)\)
• The expression is negative on \((-1, 0)\)
• The expression is positive on \((0, 1)\)
• The expression is negative on \((1, 2)\)
• The expression is positive on \((2, +\infty)\)

We include critical points where the expression equals zero (but ensure the denominator is not zero): \( x = 0 \) and \( x = 2 \). Thus, the solution is \( (-\infty, -1) \, \cup \, [0, 1) \, \cup \, (2, +\infty) \). By following these steps thoroughly, any polynomial inequality can be tackled successfully.