consider the function f(X) =`3x^(4)+4x^(3)-12x^(2)`
Y= f(X) increase in the inerval

Options

A. `(-1,0)cup(2,oo)`
B. `(-oo,0)cup(1,2)`
C. `(-2,0)cup(1,oo)`
D. none of these

Explanation

To determine the intervals where the function \( f(x) = 3x^4 + 4x^3 - 12x^2 \) is increasing, we need to follow these steps:

### Step 1: Differentiate the function
We start by finding the first derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(3x^4 + 4x^3 - 12x^2) \]

Using the power rule, we differentiate each term:

\[ f'(x) = 12x^3 + 12x^2 - 24x \]

### Step 2: Set the derivative greater than zero
For the function to be increasing, we need to find where the derivative is greater than zero:

\[ f'(x) > 0 \]

This gives us the inequality:

\[ 12x^3 + 12x^2 - 24x > 0 \]

### Step 3: Factor the derivative
We can factor out the common term \( 12x \):

\[ 12x(x^2 + x - 2) > 0 \]

Now we need to factor the quadratic \( x^2 + x - 2 \):

\[ x^2 + x - 2 = (x - 1)(x + 2) \]

So, we rewrite the inequality:

\[ 12x(x - 1)(x + 2) > 0 \]

### Step 4: Find the critical points
The critical points occur when the expression equals zero:

\[ 12x(x - 1)(x + 2) = 0 \]

Setting each factor to zero gives us:

1. \( 12x = 0 \) → \( x = 0 \)
2. \( x - 1 = 0 \) → \( x = 1 \)
3. \( x + 2 = 0 \) → \( x = -2 \)

Thus, the critical points are \( x = -2, 0, 1 \).

### Step 5: Test intervals around the critical points
We will test the sign of \( f'(x) \) in the intervals determined by the critical points:

1. \( (-\infty, -2) \)
2. \( (-2, 0) \)
3. \( (0, 1) \)
4. \( (1, \infty) \)

**Interval \( (-\infty, -2) \)**: Choose \( x = -3 \)
\[ f'(-3) = 12(-3)(-3 - 1)(-3 + 2) = 12(-3)(-4)(-1) > 0 \quad \text{(positive)} \]

**Interval \( (-2, 0) \)**: Choose \( x = -1 \)
\[ f'(-1) = 12(-1)(-1 - 1)(-1 + 2) = 12(-1)(-2)(1) > 0 \quad \text{(positive)} \]

**Interval \( (0, 1) \)**: Choose \( x = 0.5 \)
\[ f'(0.5) = 12(0.5)(0.5 - 1)(0.5 + 2) = 12(0.5)(-0.5)(2.5) < 0 \quad \text{(negative)} \]

**Interval \( (1, \infty) \)**: Choose \( x = 2 \)
\[ f'(2) = 12(2)(2 - 1)(2 + 2) = 12(2)(1)(4) > 0 \quad \text{(positive)} \]

### Step 6: Determine the intervals of increase
From our tests, we find that \( f'(x) > 0 \) in the intervals:

- \( (-\infty, -2) \)
- \( (-2, 0) \)
- \( (1, \infty) \)

Thus, the function \( f(x) \) is increasing in the intervals:

\[ (-\infty, -2) \cup (-2, 0) \cup (1, \infty) \]

### Final Answer
The function \( f(x) \) is increasing in the intervals:

\[ (-\infty, -2) \cup (-2, 0) \cup (1, \infty) \]

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