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

We are tasked with solving the inequality:

\[
x^3 - 9x \leq 0
\]

### Step 1: Factor the expression
First, factor the left-hand side of the inequality:

\[
x^3 - 9x = x(x^2 - 9)
\]

Recognize that \(x^2 - 9\) is a difference of squares, so it can be factored further:

\[
x(x^2 - 9) = x(x - 3)(x + 3)
\]

Thus, the inequality becomes:

\[
x(x - 3)(x + 3) \leq 0
\]

### Step 2: Determine critical points
The critical points are found by setting each factor equal to zero:

\[
x = 0, \quad x - 3 = 0 \implies x = 3, \quad x + 3 = 0 \implies x = -3
\]

Thus, the critical points are \(x = -3\), \(x = 0\), and \(x = 3\). 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 \(x(x - 3)(x + 3)\) in each of the following intervals: 

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

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

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

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

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

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

### Step 4: Include critical points
Since the inequality is \( \leq 0 \), we include the critical points where the expression equals zero. These occur at \(x = -3\), \(x = 0\), and \(x = 3\).

### Step 5: Write the solution in interval notation
From our sign analysis, the expression is less than or equal to zero on the intervals:

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

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

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

Would you like further details or clarifications?

### Related Questions:
1. How do you solve inequalities with higher powers like \(x^4\)?
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 (\(< 0\))?
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.

Solve the following inequality. Write the answer in interval notation. Note: If the answer includes more than one interval write the intervals separated by the union symbol, U. If needed enter ∞ as infinity and −∞ as -infinity. x^3 − 9x ≤ 0

Solve the polynomial inequality x^3 - 9x ≤ 0 using factoring and interval testing. This problem involves critical point analysis and understanding the difference of squares. Suitable for students in Grades 9-12.

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