Problem 56 - Algebra and Trigonometry

Question

Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ 16 x \leq x^{3} $$

Step-by-Step Solution

Answer

The solution in interval notation is \([-4, 0] \cup [0, 4]\).

Step 1: Bring all terms to one side

Start by rewriting the given inequality by bringing all terms to the left side. The inequality becomes: \[ x^3 - 16x \leq 0 \]

Step 2: Factor the expression

Factor the expression on the left side. Notice that you can factor out an \(x\): \[ x(x^2 - 16) \leq 0 \]Next, factor \(x^2 - 16\) as a difference of squares: \[ x(x - 4)(x + 4) \leq 0 \]

Step 3: Find the critical points

The critical points are found by setting each factor equal to zero: 1. \(x = 0\)2. \(x - 4 = 0\) which gives \(x = 4\)3. \(x + 4 = 0\) which gives \(x = -4\) So, the critical points are \(x = -4, 0, 4\).

Step 4: Test intervals between critical points

The critical points divide the real number line into intervals: 1. \((-\infty, -4)\)2. \((-4, 0)\)3. \((0, 4)\)4. \((4, \infty)\)Choose a test point from each interval and evaluate the expression \(x(x-4)(x+4)\) to see if it is less than or equal to 0. - Interval \((-\infty, -4)\): Choose \(x = -5\), \((-5)((-5)-4)((-5)+4) = -45\), which is negative.- Interval \((-4, 0)\): Choose \(x = -1\), \((-1)((-1)-4)((-1)+4) = 15\), which is positive.- Interval \((0, 4)\): Choose \(x = 1\), \((1)((1)-4)((1)+4) = -15\), which is negative.- Interval \((4, \infty)\): Choose \(x = 5\), \((5)((5)-4)((5)+4) = 45\), which is positive.

Step 5: Determine the solution set including endpoints

From the test, intervals \((-\infty, -4)\) and \((0, 4)\) result in negative values whereas the inequality requires them to be \(\leq 0\). Include critical points because the inequality allows equality (i.e., \(\leq\)): The solution includes \([-4, 0] \cup [0, 4]\).

Step 6: Graph the solution

Draw a number line and shade the intervals belonging to the solution. Include closed circles at \(x = -4\), \(x = 0\), and \(x = 4\) to denote these points are included in the solution set.

Key Concepts

Factorization

When solving a nonlinear inequality like \( 16x \leq x^3 \), factorization is essential. First, you need to rearrange the inequality: move all terms to one side so that zero is on the other side. This gives: \( x^3 - 16x \leq 0 \).

The next step is factoring. In this problem, you can first take out \( x \) as a common factor: \( x(x^2 - 16) \leq 0 \).
Then, use the difference of squares for \( x^2 - 16 \): it factors into \( (x - 4)(x + 4) \).

So, the inequality becomes \( x(x - 4)(x + 4) \leq 0 \). Breaking down an expression using factorization turns a complex problem into more manageable parts.

Understanding how to factor efficiently can simplify many algebraic processes and is often the first step in finding solutions to polynomial equations or inequalities.

Critical Points

Critical points are values of \( x \) that make each factor of a polynomial equation zero. They divide the number line into distinct intervals. For our inequality, \( x(x - 4)(x + 4) \leq 0 \), set each factor equal to zero to find critical points.

• The factor \( x = 0 \) gives the critical point \( x = 0 \).
• For \( x - 4 = 0 \), solve to find \( x = 4 \).
• For \( x + 4 = 0 \), solve to find \( x = -4 \).

Each critical point indicates where the sign of the entire expression might change. The critical points for this inequality are \( x = -4, 0, 4 \).
They are important as determining these points helps define intervals that must be tested for sign changes in the polynomial expression.

Interval Notation

Interval notation provides a concise way to describe sets of numbers, particularly solutions of inequalities. After determining the critical points, the real number line is divided into intervals. These intervals are then tested to understand where the inequality holds true.
For our inequality \( x(x - 4)(x + 4) \leq 0 \), testing the intervals derived from critical points \( -4, 0, \) and \( 4 \) helps find where the inequality is satisfied:

• From critical points, we determine the intervals: \(( -\infty, -4 )\), \(( -4, 0 )\), \(( 0, 4 )\), and \(( 4, \infty )\).
• Evaluating test points within those intervals, it turns out \( (-\infty, -4) \) and \( (0, 4) \) satisfy the inequality.

Interval notation uses brackets to show whether endpoints are included: square brackets \([\ ]\) indicate inclusion, while parentheses \((\ )\) indicate exclusion. The solution in interval notation is \([-4, 0] \cup [0, 4]\), meaning both \([-4, 0]\) and \([0, 4]\) are parts of the solution.

Graphical Representation

Visualizing the solution of a nonlinear inequality on a number line can greatly aid understanding. The number line captures where the expression holds true, providing a clear view of the solution set.
For \( x(x - 4)(x + 4) \leq 0 \), critical points \(-4, 0, \) and \( 4 \) are plotted, and the intervals \([-4, 0]\) and \([0, 4]\) are shaded to mark the solution. Closed circles at \(-4, 0,\) and \(4\) show that these endpoints are included in the solution because the inequality is \( \leq 0 \).

• Shade the section between \(-4\) and \(0\) on the line, confirming inequality conditions.
• Similarly, shade between \(0\) and \(4\).

Graphical representation is a powerful tool; seeing the problem solved visually makes the results intuitive, highlighting where the inequality holds across the number line.