Find the values of x, for which the function `f(x)= x^(3) + 6x^(2) - 36x+ 6` is monotonically decreasing.

Answer

To find the values of \( x \) for which the function \( f(x) = x^3 + 6x^2 - 36x + 6 \) is monotonically decreasing, we need to follow these steps:

### Step 1: Find the derivative of the function
To determine where the function is monotonically decreasing, we first need to compute the derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(x^3 + 6x^2 - 36x + 6) \]

Calculating the derivative term by term:

- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( 6x^2 \) is \( 12x \).
- The derivative of \( -36x \) is \( -36 \).
- The derivative of the constant \( 6 \) is \( 0 \).

Putting it all together:

\[ f'(x) = 3x^2 + 12x - 36 \]

### Step 2: Set the derivative less than zero
For the function to be monotonically decreasing, we need:

\[ f'(x) < 0 \]

This means:

\[ 3x^2 + 12x - 36 < 0 \]

### Step 3: Simplify the inequality
We can simplify this inequality by dividing all terms by \( 3 \):

\[ x^2 + 4x - 12 < 0 \]

### Step 4: Factor the quadratic expression
Next, we need to factor the quadratic expression \( x^2 + 4x - 12 \). We look for two numbers that multiply to \( -12 \) and add to \( 4 \). The numbers \( 6 \) and \( -2 \) work:

\[ x^2 + 4x - 12 = (x - 2)(x + 6) \]

### Step 5: Solve the inequality
Now we need to solve the inequality:

\[ (x - 2)(x + 6) < 0 \]

### Step 6: Determine the critical points
The critical points from the factors are:

- \( x - 2 = 0 \) gives \( x = 2 \)
- \( x + 6 = 0 \) gives \( x = -6 \)

### Step 7: Test intervals
We will test the intervals determined by the critical points \( -6 \) and \( 2 \):

1. **Interval \( (-\infty, -6) \)**: Choose \( x = -7 \)
\[ (-7 - 2)(-7 + 6) = (-9)(-1) = 9 > 0 \]

2. **Interval \( (-6, 2) \)**: Choose \( x = 0 \)
\[ (0 - 2)(0 + 6) = (-2)(6) = -12 < 0 \]

3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \)
\[ (3 - 2)(3 + 6) = (1)(9) = 9 > 0 \]

### Step 8: Conclusion
The function \( f(x) \) is monotonically decreasing in the interval where the product is negative:

\[ x \in (-6, 2) \]

Thus, the values of \( x \) for which the function \( f(x) \) is monotonically decreasing are:

\[ \boxed{(-6, 2)} \]