(3x-2) (x-4) (3-2x)<0
To solve the inequality \((3x-2)(x-4)(3-2x) < 0\), we will first identify the points where the expression is equal to zero, as these points will help us determine the intervals for testing.
Find the zeros:
Set each factor equal to zero:
- \(3x - 2 = 0\) → \(x = \frac{2}{3}\)
- \(x - 4 = 0\) → \(x = 4\)
- \(3 - 2x = 0\) → \(x = \frac{3}{2}\)
Now, we have the critical points:
- \(x = \frac{2}{3}\)
- \(x = \frac{3}{2}\)
- \(x = 4\)
Determine intervals:
The critical points divide the number line into the following intervals:
- \((-∞, \frac{2}{3})\)
- \((\frac{2}{3}, \frac{3}{2})\)
- \((\frac{3}{2}, 4)\)
- \((4, ∞)\)
Test the intervals:
We will pick a test point from each interval to determine whether the product is negative or positive.
Interval \((-∞, \frac{2}{3})\): Let \(x = 0\) \[ (3(0) - 2)(0 - 4)(3 - 2(0)) = (-2)(-4)(3) = 24 > 0 \]
Interval \((\frac{2}{3}, \frac{3}{2})\): Let \(x = 1\) \[ (3(1) - 2)(1 - 4)(3 - 2(1)) = (1)(-3)(1) = -3 < 0 \]
Interval \((\frac{3}{2}, 4)\): Let \(x = 2\) \[ (3(2) - 2)(2 - 4)(3 - 2(2)) = (6 - 2)(-2)(3 - 4) = (4)(-2)(-1) = 8 > 0 \]
Interval \((4, ∞)\): Let \(x = 5\) \[ (3(5) - 2)(5 - 4)(3 - 2(5)) = (15 - 2)(1)(3 - 10) = (13)(1)(-7) = -91 < 0 \]
Determine where the expression is negative:
From our tests:
- The product is positive in \((-∞, \frac{2}{3})\) and \((\frac{3}{2}, 4)\).
- The product is negative in \((\frac{2}{3}, \frac{3}{2})\) and \((4, ∞)\).
Combine intervals:
The solution to the inequality \((3x - 2)(x - 4)(3 - 2x) < 0\) is:
\[ \boxed{(\frac{2}{3}, \frac{3}{2}) \cup (4, ∞)} \]
Check inclusivity: Since we are looking for where the product is strictly less than zero, we do not include the zeros (\(\frac{2}{3}, \frac{3}{2}, 4\)) in our final answer.
