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 \( g'(x) \)**:
\[ g'(x) = \frac{d}{dx}\left(2f\left(\frac{x}{2}\right) + f(2 - x)\right) \]
Using the chain rule, we differentiate each term:
\[ g'(x) = 2 \cdot f'\left(\frac{x}{2}\right) \cdot \frac{1}{2} + f'(2 - x) \cdot (-1) \]
Simplifying this gives:
\[ g'(x) = f'\left(\frac{x}{2}\right) - f'(2 - x) \]

2. **Determine when \( g'(x) > 0 \)**:
We want to find the values of \( x \) such that:
\[ f'\left(\frac{x}{2}\right) > f'(2 - x) \]

3. **Analyze the behavior of \( f' \)**:
Since \( f''(x) < 0 \) for \( x \in (0, 2) \), this means that \( f'(x) \) is a decreasing function. Therefore, if \( x_1 < x_2 \), then:
\[ f'(x_1) > f'(x_2) \]

4. **Set up the inequality**:
Let \( a = \frac{x}{2} \) and \( b = 2 - x \). We need to find when:
\[ f'(a) > f'(b) \]
Given that \( f' \) is decreasing, this inequality implies:
\[ a < b \]
Substituting back for \( a \) and \( b \):
\[ \frac{x}{2} < 2 - x \]

5. **Solve the inequality**:
Rearranging the inequality:
\[ \frac{x}{2} + x < 2 \]
\[ \frac{3x}{2} < 2 \]
\[ 3x < 4 \]
\[ x < \frac{4}{3} \]

6. **Consider the domain**:
Since \( x \) must also be in the interval \( (0, 2) \), we combine the results:
\[ 0 < x < \frac{4}{3} \]

7. **Final interval**:
Therefore, the interval in which \( g(x) \) is increasing is:
\[ \boxed{(0, \frac{4}{3})} \]