Let `f''(x) gt 0 AA x in R and ` let `g(x)=f(x)+f(2-x)` then interval of x for which g(x) is increasing is

Options

A. `(1,oo)`
B. R
C. `[-1,1]`
D. `[-2,oo)`

Answer

To solve the problem, we need to find the interval of \( x \) for which the function \( g(x) = f(x) + f(2 - x) \) is increasing. Given that \( f''(x) > 0 \) for all \( x \in \mathbb{R} \), this indicates that \( f(x) \) is a convex function.

### Step-by-Step Solution:

1. **Define the function**:
\[ g(x) = f(x) + f(2 - x) \]

2. **Differentiate \( g(x) \)**:
To find when \( g(x) \) is increasing, we need to compute the first derivative \( g'(x) \):
\[ g'(x) = f'(x) + \frac{d}{dx}[f(2 - x)] \]
Using the chain rule, we have:
\[ \frac{d}{dx}[f(2 - x)] = f'(2 - x) \cdot (-1) = -f'(2 - x) \]
Therefore,
\[ g'(x) = f'(x) - f'(2 - x) \]

3. **Set the derivative greater than zero**:
For \( g(x) \) to be increasing, we need:
\[ g'(x) > 0 \implies f'(x) - f'(2 - x) > 0 \implies f'(x) > f'(2 - x) \]

4. **Analyze the inequality**:
The inequality \( f'(x) > f'(2 - x) \) suggests that the slope of \( f \) at \( x \) is greater than the slope of \( f \) at \( 2 - x \).

5. **Consider the implications of convexity**:
Since \( f''(x) > 0 \), \( f'(x) \) is an increasing function. Therefore, if \( x > 2 - x \), then \( x > 1 \) implies \( f'(x) > f'(2 - x) \).

6. **Solve the inequality**:
\[ x > 2 - x \implies 2x > 2 \implies x > 1 \]

7. **Determine the interval**:
Thus, \( g(x) \) is increasing for:
\[ x \in (1, \infty) \]

### Final Answer:
The interval of \( x \) for which \( g(x) \) is increasing is \( (1, \infty) \).