To solve the inequality \(\frac{(x-1)(x-2)(x-3)^2}{(x-4)^2(x-5)^3} < 0\), we will follow these steps:
### Step 1: Identify the critical points
The critical points occur when the numerator or denominator is equal to zero.
- **Numerator**:
- \(x - 1 = 0 \Rightarrow x = 1\)
- \(x - 2 = 0 \Rightarrow x = 2\)
- \((x - 3)^2 = 0 \Rightarrow x = 3\) (note that this is a repeated root)
- **Denominator**:
- \((x - 4)^2 = 0 \Rightarrow x = 4\) (note that this is a repeated root)
- \((x - 5)^3 = 0 \Rightarrow x = 5\) (note that this is a repeated root)
Thus, the critical points are \(x = 1, 2, 3, 4, 5\).
### Step 2: Plot the critical points on a number line
We will place the critical points on a number line:
- Open intervals: \( (-\infty, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, \infty) \)
### Step 3: Test the intervals
We will test points from each interval to determine the sign of the expression.
1. **Interval \((- \infty, 1)\)**: Choose \(x = 0\)
\[
\frac{(0-1)(0-2)(0-3)^2}{(0-4)^2(0-5)^3} = \frac{(-1)(-2)(9)}{(16)(-125)} = \frac{18}{-2000} < 0
\]
2. **Interval \((1, 2)\)**: Choose \(x = 1.5\)
\[
\frac{(1.5-1)(1.5-2)(1.5-3)^2}{(1.5-4)^2(1.5-5)^3} = \frac{(0.5)(-0.5)(2.25)}{(-2.5)^2(-3.5)^3} > 0
\]
3. **Interval \((2, 3)\)**: Choose \(x = 2.5\)
\[
\frac{(2.5-1)(2.5-2)(2.5-3)^2}{(2.5-4)^2(2.5-5)^3} = \frac{(1.5)(0.5)(0.25)}{(-1.5)^2(-2.5)^3} < 0
\]
4. **Interval \((3, 4)\)**: Choose \(x = 3.5\)
\[
\frac{(3.5-1)(3.5-2)(3.5-3)^2}{(3.5-4)^2(3.5-5)^3} = \frac{(2.5)(1.5)(0.25)}{(-0.5)^2(-1.5)^3} < 0
\]
5. **Interval \((4, 5)\)**: Choose \(x = 4.5\)
\[
\frac{(4.5-1)(4.5-2)(4.5-3)^2}{(4.5-4)^2(4.5-5)^3} = \frac{(3.5)(2.5)(2.25)}{(0.5)^2(-0.5)^3} > 0
\]
6. **Interval \((5, \infty)\)**: Choose \(x = 6\)
\[
\frac{(6-1)(6-2)(6-3)^2}{(6-4)^2(6-5)^3} = \frac{(5)(4)(9)}{(2)(1)} > 0
\]
### Step 4: Combine the results
From our tests, the function is negative in the intervals:
- \((- \infty, 1)\)
- \((2, 3)\)
- \((3, 4)\)
### Step 5: Write the solution set
Since the inequality is strict (\(< 0\)), we exclude the critical points. Thus, the solution set is:
\[
(-\infty, 1) \cup (2, 3) \cup (3, 4)
\]
