Problem 40 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-40: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ x^{4} \geq x^{2} $$

Step-by-Step Solution

Answer

The solution in terms of intervals is: \((-\infty, -1] \cup [0, 1] \cup (1, \infty)\).

Step 1: Simplify the inequality

Start by bringing all terms to one side of the inequality: \[ x^4 - x^2 \geq 0 \] Factor out the common term: \[ x^2(x^2 - 1) \geq 0 \]

Step 2: Further factorize

Notice that \( x^2 - 1 \) can be factored as a difference of squares: \[ x^2 - 1 = (x - 1)(x + 1) \] Therefore, the inequality becomes: \[ x^2(x - 1)(x + 1) \geq 0 \]

Step 3: Find critical points

The critical points occur where each factor is equal to zero: 1. \( x^2 = 0 \) gives \( x = 0 \) 2. \( x - 1 = 0 \) gives \( x = 1 \) 3. \( x + 1 = 0 \) gives \( x = -1 \) These critical points divide the real number line into intervals.

Step 4: Test intervals

Choose test points from each of the intervals created by the critical points \(-\infty, -1, 0, 1, \infty \) and determine where the inequality holds:1. Test \( x = -2 \) in \((-\infty, -1)\): \((-2)^2(-2-1)(-2+1) = 4(-3)(-1) = 12 \geq 0\). True.2. Test \( x = -0.5 \) in \((-1, 0)\): \((-0.5)^2(-0.5-1)(-0.5+1) = 0.25(-1.5)(0.5) = -0.1875 gtr 0\). False.3. Test \( x = 0.5 \) in \((0, 1)\): \((0.5)^2(0.5-1)(0.5+1) = 0.25(-0.5)(1.5) = -0.1875 gtr 0\). False.4. Test \( x = 2 \) in \((1, \infty)\): \(2^2(2-1)(2+1) = 4(1)(3) = 12 \geq 0\). True.The inequality holds in the intervals \((-\infty, -1) \cup [0, 0] \cup (1, \infty)\).

Step 5: Consider equality points

Evaluate the inequality at the critical points:1. For \( x = -1 \), \( x^2(x - 1)(x + 1) = 1(0)(-2) = 0 \geq 0 \). True.2. For \( x = 0 \), \( x^2(x - 1)(x + 1) = 0 \). True.3. For \( x = 1 \), \( x^2(x - 1)(x + 1) = 1(0)(2) = 0 \geq 0 \). True.Since the inequality holds at these points, they are included in the solution set.

Key Concepts

Critical Points

When solving inequalities, identifying critical points is a crucial step. These points are where the expression changes sign. In other words, they help us determine where our inequality might switch from being true to false or vice versa.

Finding the critical points involves setting each factor of your expression to zero:

• If you have an expression like \(x^2(x-1)(x+1)\), you find the critical points by solving \(x^2 = 0\), \(x - 1 = 0\), and \(x + 1 = 0\).
• This will give you the values of \(x\) where the sign of the entire expression might change. Here, the critical points are \(x = 0\), \(x = 1\), and \(x = -1\).

These points divide the real number line into intervals. By testing points within these intervals, we figure out on which intervals the inequality holds true.

Factorization

Factorization is all about breaking down expressions into products of simpler terms. It's like dissecting the original polynomial into its foundational building blocks.

In our example, the goal was to factor \(x^4 - x^2\). The first step was to factor out the common term \(x^2\), which simplifies the expression to \(x^2(x^2 - 1)\).

Next, we further factor \(x^2 - 1\) using the difference of squares formula. This results in the factors \((x-1)(x+1)\). The expression now reads \(x^2(x-1)(x+1)\).

Factorization helps us solve the inequality because it exposes the critical points—where each factor equals zero. This tells us key transition points for the inequality's solutions.

Interval Notation

Interval notation is a concise way to describe sets of numbers, especially when we are dealing with inequalities. Instead of listing individual numbers, we describe a continuous block of numbers.

How do you express intervals?

• When the inequality holds in an interval, we use brackets: \([\text{start}, \text{end}])\).
• Use a parenthesis \(()\) to indicate that an endpoint is not included.

For example:

• \((-\infty, -1) \cup [0, 0] \cup (1, \infty)\) means:
  • The interval from \(-\infty\) to \(-1\), not including \(-1\).
  • The single number \(0\) (included).
  • The interval from \(1\) to \(\infty\), not including \(1\).

The use of \(\infty\) or \(-\infty\) always employs parentheses since infinity is not a number but a concept.

Real Number Line

The real number line is a visual way to represent all possible values that a real number can take. Think of it as an endless line stretching in both directions that includes every number you could think of, even decimals and irrational numbers.

In the context of inequalities:

• Critical points are marked on this line, helping to partition it into segments or intervals.
• You analyze each interval to discover where the inequality holds true.

Picture the real number line and place the critical points of \(-1, 0,\) and \(1\) on it. It divides the line into:

• \((-\infty, -1)\)
• \((-1, 0)\)
• \((0, 1)\)
• and \((1, \infty)\)

Within these intervals, we test values to determine where the inequality is true, using our understanding of each segment marked by critical points.