If `g(x) =2f(2x^3-3x^2)+f(6x^2-4x^3-3) AA x in R` and `f''(x) gt 0 AA x in R` then g(x) is increasing in the interval

Options

A. `(-oo,-(1)/(2))uu(0,1)`
B. `(-(1)/(2),0)uu(1,oo)`
C. `(0,oo)`
D. `(-oo,1)`

Explanation

To determine the intervals where the function \( g(x) = 2f(2x^3 - 3x^2) + f(6x^2 - 4x^3 - 3) \) is increasing, we need to analyze the derivative \( g'(x) \) and the properties of the function \( f \).

### Step 1: Find the derivative \( g'(x) \)

Given:
\[ g(x) = 2f(2x^3 - 3x^2) + f(6x^2 - 4x^3 - 3) \]

Using the chain rule, we differentiate \( g(x) \):

\[ g'(x) = 2f'(2x^3 - 3x^2) \cdot (6x^2 - 6x) + f'(6x^2 - 4x^3 - 3) \cdot (12x - 12x^2) \]

### Step 2: Simplify \( g'(x) \)

This can be simplified as follows:

\[ g'(x) = 2f'(2x^3 - 3x^2)(6x^2 - 6x) + f'(6x^2 - 4x^3 - 3)(12x - 12x^2) \]

Factoring out common terms:

\[ g'(x) = 12x(1 - x)(f'(2x^3 - 3x^2) - f'(6x^2 - 4x^3 - 3)) \]

### Step 3: Determine when \( g'(x) > 0 \)

For \( g(x) \) to be increasing, we need \( g'(x) > 0 \):

\[ 12x(1 - x)(f'(2x^3 - 3x^2) - f'(6x^2 - 4x^3 - 3) > 0 \]

### Step 4: Analyze the factors

1. **Factor \( 12x \)**: This is positive when \( x > 0 \).
2. **Factor \( (1 - x) \)**: This is positive when \( x < 1 \).
3. **Factor \( (f'(2x^3 - 3x^2) - f'(6x^2 - 4x^3 - 3) \)**: Since \( f''(x) > 0 \), \( f'(x) \) is increasing. Thus, we need to analyze when \( 2x^3 - 3x^2 < 6x^2 - 4x^3 - 3 \).

### Step 5: Solve the inequality

Rearranging gives:

\[ 2x^3 + 4x^3 - 3x^2 + 3 < 0 \]

This simplifies to:

\[ 6x^3 - 3x^2 + 3 < 0 \]

### Step 6: Find critical points

To find the intervals where \( g'(x) > 0 \), we need to find the roots of \( 6x^3 - 3x^2 + 3 = 0 \). Testing values will help us find where the function changes sign.

### Step 7: Conclusion about intervals

From the analysis, we find that:

- For \( x < 0 \), \( g'(x) > 0 \).
- For \( 0 < x < 1 \), \( g'(x) < 0 \).
- For \( x > 1 \), \( g'(x) > 0 \).

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

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

### Final Answer

The intervals where \( g(x) \) is increasing are:

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