Let `g(x) = 2 f(x/2)+ f(2-x) and f''(x) lt 0 forall x in (0,2)`. Then calculate the interval in which g(x) is increasing.

Answer

To solve the problem, we need to determine the interval in which the function \( g(x) = 2f\left(\frac{x}{2}\right) + f(2-x) \) is increasing, given that \( f''(x) < 0 \) for all \( x \) in the interval \( (0, 2) \).

### Step-by-Step Solution:

1. **Find the derivative \( g'(x) \)**:
We start by differentiating \( g(x) \):
\[ g'(x) = 2 \cdot f'\left(\frac{x}{2}\right) \cdot \frac{1}{2} + f'(2-x) \cdot (-1) \]
Simplifying this, we get:
\[ g'(x) = f'\left(\frac{x}{2}\right) - f'(2-x) \]

2. **Determine when \( g'(x) > 0 \)**:
For \( g(x) \) to be increasing, we need:
\[ g'(x) > 0 \implies f'\left(\frac{x}{2}\right) > f'(2-x) \]

3. **Analyze the behavior of \( f' \)**:
Given that \( f''(x) < 0 \) for all \( x \in (0, 2) \), we know that \( f' \) is a decreasing function in this interval. Therefore, if \( a < b \), then \( f'(a) > f'(b) \).

4. **Set up the inequality**:
We need to find the values of \( x \) such that:
\[ \frac{x}{2} < 2 - x \]
This simplifies to:
\[ x < 4 - 2x \]
Rearranging gives:
\[ 3x < 4 \implies x < \frac{4}{3} \]

5. **Determine the interval**:
Since \( x \) must also be in the interval \( (0, 2) \) (as \( g(x) \) is defined for \( x \) in this range), we conclude that:
\[ g(x) \text{ is increasing on } \left(0, \frac{4}{3}\right) \]

### Final Answer:
The function \( g(x) \) is increasing on the interval \( \left(0, \frac{4}{3}\right) \).