(4x-1)(x^2-4)<0

To solve the inequality \((4x-1)(x^2-4)<0\), we can first find the points where the expression equals zero. This will help us determine the intervals to test for the inequality.

1. Set the expression equal to zero:

\[(4x - 1)(x^2 - 4) = 0\]

This gives us two factors to consider:

• \(4x - 1 = 0\) which implies \(x = \frac{1}{4}\)
• \(x^2 - 4 = 0\) which factors to \((x - 2)(x + 2) = 0\) leading to the solutions \(x = 2\) and \(x = -2\)

Now, we have the critical points:

• \(x = -2\)
• \(x = \frac{1}{4}\)
• \(x = 2\)

2. Next, we will use these critical points to divide the real line into intervals:

• \((- \infty, -2)\)
• \((-2, \frac{1}{4})\)
• \((\frac{1}{4}, 2)\)
• \((2, \infty)\)

3. We will test a point from each interval to determine where the product \((4x - 1)(x^2 - 4)\) is negative.

• For the interval \((- \infty, -2)\), let's test \(x = -3\): \((4(-3) - 1)((-3)^2 - 4) = (-12 - 1)(9 - 4) = (-13)(5) < 0\)

• For the interval \((-2, \frac{1}{4})\), let's test \(x = 0\): \((4(0) - 1)((0)^2 - 4) = (-1)(-4) = 4 > 0\)

• For the interval \((\frac{1}{4}, 2)\), let's test \(x = 1\): \((4(1) - 1)((1)^2 - 4) = (4 - 1)(1 - 4) = (3)(-3) < 0\)

• For the interval \((2, \infty)\), let's test \(x = 3\): \((4(3) - 1)((3)^2 - 4) = (12 - 1)(9 - 4) = (11)(5) > 0\)

4. Now, we summarize the signs of the product in each interval:

• \((- \infty, -2)\): negative
• \((-2, \frac{1}{4})\): positive
• \((\frac{1}{4}, 2)\): negative
• \((2, \infty)\): positive

5. The inequality \((4x - 1)(x^2 - 4) < 0\) holds for the intervals where the product is negative:

• \((- \infty, -2)\)
• \((\frac{1}{4}, 2)\)

Thus, the solution to the inequality is:

\[ (-\infty, -2) \cup \left(\frac{1}{4}, 2\right) \]