Problem 33 - Calculus with Concepts in Calculus

Question

Solve the inequality. $$ \frac{4 x\left(x^{2}-6\right)}{x^{2}-4}<0 $$

Step-by-Step Solution

Answer

The solution is \((-2, 0) \cup (0, 2)\).

Step 1: Identify the Critical Points

The inequality is \( \frac{4x(x^2 - 6)}{x^2 - 4} < 0 \). To find the critical points, set 1) The numerator equal to zero: \[ 4x(x^2 - 6) = 0 \] This gives \( x = 0 \) and \( x^2 = 6 \) which implies \( x = \pm \sqrt{6} \).2) The denominator equal to zero: \[ x^2 - 4 = 0 \] This gives \( x = \pm 2 \). Thus, the critical points are \( x = -\sqrt{6}, -2, 0, 2, \sqrt{6} \).

Step 2: Determine Intervals

The critical points divide the real number line into intervals, which are:1. \(( -\infty, -\sqrt{6} ) \)2. \(( -\sqrt{6}, -2 ) \)3. \(( -2, 0 ) \)4. \(( 0, 2 ) \)5. \(( 2, \sqrt{6} ) \)6. \(( \sqrt{6}, \infty ) \)

Step 3: Test Each Interval

Select a test point in each interval and substitute into the inequality:- Interval \(( -\infty, -\sqrt{6} )\): Test with \( x = -3 \).- Interval \(( -\sqrt{6}, -2 )\): Test with \( x = -3 \ 2 \).- Interval \(( -2, 0 )\): Test with \( x = -1 \).- Interval \(( 0, 2 )\): Test with \( x = 1 \).- Interval \(( 2, \sqrt{6} )\): Test with \( x = 3 \ 2 \).- Interval \(( \sqrt{6}, \infty )\): Test with \( x = 3 \).\[- (-3) \Rightarrow \frac{4(-3)((-3)^2 - 6)}{(-3)^2 - 4} > 0 - (-\frac{5}{2}) \Rightarrow \frac{4(-2.5)((-2.5)^2 - 6)}{(-2.5)^2 - 4} > 0- (-1) \Rightarrow \frac{4(-1)((-1)^2 - 6)}{(-1)^2 - 4} < 0- (1) \Rightarrow \frac{4(1)((1)^2 - 6)}{(1)^2 - 4} < 0- (\frac{5}{2}) \Rightarrow \frac{4(2.5)((2.5)^2 - 6)}{(2.5)^2 - 4} > 0- (3) \Rightarrow \frac{4(3)((3)^2 - 6)}{(3)^2 - 4} > 0\]

Step 4: Determine Solution Set

The inequality holds true in intervals where the expression evaluates to a negative value. From the tests:- Interval \((-2, 0)\) is true.- Interval \((0, 2)\) is true as well.Thus, the solution set is \((-2, 0) \cup (0, 2)\).

Key Concepts

Critical Points

Critical points are key to understanding where an inequality changes its nature. The critical points of \[ \frac{4x(x^2 - 6)}{x^2 - 4} < 0 \]are the values of \(x\) where either the numerator or the denominator equals zero. These are the points where the inequality could change from negative to positive or vice versa.
To find these, we:

• Set the numerator equal to zero.
• Set the denominator equal to zero.

For this specific inequality:

• The numerator, \(4x(x^2 - 6) = 0\), leads to \(x = 0\) or \(x^2 = 6\). Thus, \(x = \pm \sqrt{6}\).
• The denominator, \(x^2 - 4 = 0\), simplifies to \(x = \pm 2\).

Together, these values form our critical points: \(-\sqrt{6}, -2, 0, 2, \sqrt{6}\). These points will help us to divide the real number line into intervals for further testing.

Interval Testing

After determining the critical points, the next step is to use them to divide the number line into distinct intervals. Interval testing involves selecting test points from each of these intervals to evaluate the inequality.
The intervals for the given inequality are:

• \((-\infty, -\sqrt{6})\)
• \((-\sqrt{6}, -2)\)
• \((-2, 0)\)
• \((0, 2)\)
• \((2, \sqrt{6})\)
• \((\sqrt{6}, \infty)\)

Choose a test point from each interval and substitute it back into the original inequality. Check whether the resulting expression is positive or negative. This evaluation will help to determine where the inequality is satisfied or not.
For example:

• Testing with \(x = -3\) in \((-\infty, -\sqrt{6})\), results in a positive value.
• Testing with \(x = 1\) in \((0, 2)\), results in a negative value, indicating the inequality is true here.

Inequality Solution Set

The solution set of an inequality refers to the range of values for which the inequality holds true. Once we perform interval testing, it results in particular intervals being identified where the inequality expression is less than zero. These form the inequality solution set.
In our example, after testing each interval:

• The inequality is satisfied in \((-2, 0)\).
• Additionally, it is satisfied in \((0, 2)\).

Therefore, the solution set for the inequality \(\frac{4x(x^2 - 6)}{x^2 - 4} < 0\) is the union of these intervals, represented as \((-2, 0) \cup (0, 2)\). This embraces the notion of connecting the valid intervals where the inequality remains true.

Numerator and Denominator Analysis

Analyzing the numerator and denominator separately allows us to understand where changes in sign might occur. For any rational inequality:

• The numerator being zero can result in a zero value for the expression.
• The denominator being zero indicates potential undefined points for the expression.

Let's address this in our specific problem:

• The numerator, \(4x(x^2 - 6)\), becomes zero at \(x = 0\) and \(x^2 = 6\) (i.e., \(x = \pm \sqrt{6}\)). These points are candidate sites for sign changes, where the value transitions from positive to negative or the converse.
• The denominator, \(x^2 - 4\), leads to zero at \(x = \pm 2\). Here, the expression is undefined; however, these are also crucial in identifying the intervals.

Consideration of where these components of our rational expression change helps in systematically dissecting the inequality and locating solution intervals. These points of zero help map out the calculus of the inequality across the real number line.