Solve `x(x+2)^2(x-1)^5(2x-3)(x-3)^4geq0.`

Options

A. `x``in[0,1]uu[3/2,oo)`
B. `x``in[3/2,oo)`
C. `x``in[0,1]`
D. none of these

Answer

To solve the inequality \( x(x+2)^2(x-1)^5(2x-3)(x-3)^4 \geq 0 \), we will follow these steps:

### Step 1: Identify the expression
Let \( E = x(x+2)^2(x-1)^5(2x-3)(x-3)^4 \).

### Step 2: Find the roots of the expression
We need to find the values of \( x \) for which \( E = 0 \). This occurs when any factor of \( E \) is equal to zero.

1. \( x = 0 \)
2. \( x + 2 = 0 \) → \( x = -2 \)
3. \( x - 1 = 0 \) → \( x = 1 \)
4. \( 2x - 3 = 0 \) → \( x = \frac{3}{2} \)
5. \( x - 3 = 0 \) → \( x = 3 \)

### Step 3: List the critical points
The critical points from the above calculations are:
- \( x = -2 \)
- \( x = 0 \)
- \( x = 1 \)
- \( x = \frac{3}{2} \)
- \( x = 3 \)

### Step 4: Determine the sign of \( E \) in each interval
We will analyze the sign of \( E \) in the intervals defined by these critical points. The intervals are:
1. \( (-\infty, -2) \)
2. \( (-2, 0) \)
3. \( (0, 1) \)
4. \( (1, \frac{3}{2}) \)
5. \( (\frac{3}{2}, 3) \)
6. \( (3, \infty) \)

### Step 5: Test the sign in each interval
1. **Interval \( (-\infty, -2) \)**: Choose \( x = -3 \)
- \( E = (-3)(-1)^2(-4)^5(-9)(-6)^4 \) → Positive
2. **Interval \( (-2, 0) \)**: Choose \( x = -1 \)
- \( E = (-1)(1)^2(-2)^5(-5)(-4)^4 \) → Negative
3. **Interval \( (0, 1) \)**: Choose \( x = \frac{1}{2} \)
- \( E = \left(\frac{1}{2}\right)(\frac{5}{2})^2\left(-\frac{1}{2}\right)^5(-2)(-\frac{5}{2})^4 \) → Negative
4. **Interval \( (1, \frac{3}{2}) \)**: Choose \( x = 1.2 \)
- \( E = (1.2)(3.2)^2(0.2)^5(-0.6)(-1.8)^4 \) → Positive
5. **Interval \( (\frac{3}{2}, 3) \)**: Choose \( x = 2 \)
- \( E = (2)(4)^2(1)^5(1)(-1)^4 \) → Positive
6. **Interval \( (3, \infty) \)**: Choose \( x = 4 \)
- \( E = (4)(6)^2(3)^5(5)(1)^4 \) → Positive

### Step 6: Summary of signs
- \( (-\infty, -2) \): Positive
- \( (-2, 0) \): Negative
- \( (0, 1) \): Negative
- \( (1, \frac{3}{2}) \): Positive
- \( (\frac{3}{2}, 3) \): Positive
- \( (3, \infty) \): Positive

### Step 7: Include the critical points
Since we need \( E \geq 0 \), we include the points where \( E = 0 \):
- At \( x = -2 \): \( E = 0 \)
- At \( x = 0 \): \( E = 0 \)
- At \( x = 1 \): \( E = 0 \)
- At \( x = \frac{3}{2} \): \( E = 0 \)
- At \( x = 3 \): \( E = 0 \)

### Step 8: Final solution
The solution to the inequality \( E \geq 0 \) is:
\[ x \in (-\infty, -2] \cup [1, \frac{3}{2}] \cup [3, \infty) \]