Determine the intervals in which the following functions are strictly increasing or strictly decreasing :
`f(x)=4x^(3)-6x^(2)-72x+30`.

Answer

To determine the intervals in which the function \( f(x) = 4x^3 - 6x^2 - 72x + 30 \) is strictly increasing or strictly decreasing, we need to follow these steps:

### Step 1: Find the derivative of the function
To analyze the behavior of the function, we first find its derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(4x^3 - 6x^2 - 72x + 30) \]
Using the power rule, we differentiate each term:

\[ f'(x) = 12x^2 - 12x - 72 \]

### Step 2: Set the derivative greater than zero for increasing intervals
For the function to be strictly increasing, we need to find where \( f'(x) > 0 \).

\[ 12x^2 - 12x - 72 > 0 \]

### Step 3: Simplify the inequality
We can simplify the inequality by dividing everything by 12:

\[ x^2 - x - 6 > 0 \]

### Step 4: Factor the quadratic expression
Next, we factor the quadratic:

\[ x^2 - x - 6 = (x - 3)(x + 2) \]

Thus, we need to solve the inequality:

\[ (x - 3)(x + 2) > 0 \]

### Step 5: Determine the critical points
The critical points from the factors are \( x = 3 \) and \( x = -2 \). We will use these points to test the intervals.

### Step 6: Test the intervals
The critical points divide the number line into three intervals:
1. \( (-\infty, -2) \)
2. \( (-2, 3) \)
3. \( (3, \infty) \)

We will test each interval to see where the product \( (x - 3)(x + 2) \) is positive.

- **Interval 1: \( (-\infty, -2) \)**
Choose \( x = -3 \):
\[ (-3 - 3)(-3 + 2) = (-6)(-1) = 6 > 0 \]
Thus, \( f(x) \) is increasing in this interval.

- **Interval 2: \( (-2, 3) \)**
Choose \( x = 0 \):
\[ (0 - 3)(0 + 2) = (-3)(2) = -6 < 0 \]
Thus, \( f(x) \) is decreasing in this interval.

- **Interval 3: \( (3, \infty) \)**
Choose \( x = 4 \):
\[ (4 - 3)(4 + 2) = (1)(6) = 6 > 0 \]
Thus, \( f(x) \) is increasing in this interval.

### Step 7: Summarize the results
From our tests, we conclude:
- The function \( f(x) \) is strictly increasing on the intervals \( (-\infty, -2) \) and \( (3, \infty) \).
- The function \( f(x) \) is strictly decreasing on the interval \( (-2, 3) \).

### Final Answer
- **Strictly Increasing:** \( (-\infty, -2) \cup (3, \infty) \)
- **Strictly Decreasing:** \( (-2, 3) \)