Math Problem Statement

Solve the following inequality. Write the answerinintervalnotation.Note: Iftheanswerincludesmore than one interval write the intervals separated by the ”union” symbol, U. If needed enter ∞ as infinity and −∞ as -infinity . x3 − 9x ≤ 0

Solution

We are tasked with solving the inequality:

x39x0x^3 - 9x \leq 0

Step 1: Factor the expression

First, factor the left-hand side of the inequality:

x39x=x(x29)x^3 - 9x = x(x^2 - 9)

Recognize that x29x^2 - 9 is a difference of squares, so it can be factored further:

x(x29)=x(x3)(x+3)x(x^2 - 9) = x(x - 3)(x + 3)

Thus, the inequality becomes:

x(x3)(x+3)0x(x - 3)(x + 3) \leq 0

Step 2: Determine critical points

The critical points are found by setting each factor equal to zero:

x=0,x3=0    x=3,x+3=0    x=3x = 0, \quad x - 3 = 0 \implies x = 3, \quad x + 3 = 0 \implies x = -3

Thus, the critical points are x=3x = -3, x=0x = 0, and x=3x = 3. These points divide the real line into intervals, which we will now test.

Step 3: Test intervals

We need to test the sign of the expression x(x3)(x+3)x(x - 3)(x + 3) in each of the following intervals:

  • (,3)(-\infty, -3)
  • (3,0)(-3, 0)
  • (0,3)(0, 3)
  • (3,)(3, \infty)

Let's choose a test point in each interval and determine the sign of the product:

  1. For x=4x = -4 (in (,3)(-\infty, -3)): (4)((4)3)((4)+3)=(4)(7)(1)=28(negative)(-4)((-4) - 3)((-4) + 3) = (-4)(-7)(-1) = -28 \quad (\text{negative}) So, the expression is negative in (,3)(-∞, -3).

  2. For x=1x = -1 (in (3,0)(-3, 0)): (1)((1)3)((1)+3)=(1)(4)(2)=8(positive)(-1)((-1) - 3)((-1) + 3) = (-1)(-4)(2) = 8 \quad (\text{positive}) So, the expression is positive in (3,0)(-3, 0).

  3. For x=1x = 1 (in (0,3)(0, 3)): (1)((1)3)((1)+3)=(1)(2)(4)=8(negative)(1)((1) - 3)((1) + 3) = (1)(-2)(4) = -8 \quad (\text{negative}) So, the expression is negative in (0,3)(0, 3).

  4. For x=4x = 4 (in (3,)(3, \infty)): (4)((4)3)((4)+3)=(4)(1)(7)=28(positive)(4)((4) - 3)((4) + 3) = (4)(1)(7) = 28 \quad (\text{positive}) So, the expression is positive in (3,)(3, ∞).

Step 4: Include critical points

Since the inequality is 0\leq 0, we include the critical points where the expression equals zero. These occur at x=3x = -3, x=0x = 0, and x=3x = 3.

Step 5: Write the solution in interval notation

From our sign analysis, the expression is less than or equal to zero on the intervals:

(,3][0,3](-\infty, -3] \cup [0, 3]

Thus, the solution to the inequality x39x0x^3 - 9x \leq 0 in interval notation is:

(,3][0,3]\boxed{(-\infty, -3] \cup [0, 3]}

Would you like further details or clarifications?

Related Questions:

  1. How do you solve inequalities with higher powers like x4x^4?
  2. What is the graphical interpretation of the solution for this inequality?
  3. How do you handle inequalities involving rational expressions?
  4. How would the solution change if the inequality was strict (<0< 0)?
  5. Can this factorization technique be applied to solve polynomial equations of degree higher than 3?

Tip:

When solving inequalities, always test the sign of the expression in the intervals defined by the critical points to determine where the inequality holds.

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Math Problem Analysis

Mathematical Concepts

Algebra
Polynomial Inequalities
Factoring
Critical Points

Formulas

Factorization: x^3 - 9x = x(x - 3)(x + 3)

Theorems

Difference of Squares
Sign Testing for Inequalities

Suitable Grade Level

Grades 9-12