Let `f''(x) gt 0 AA x in R and g(x)=f(2-x)+f(4+x).` Then `g(x)` is increasing in

Options

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

Answer

To determine the interval in which the function \( g(x) = f(2 - x) + f(4 + x) \) is increasing, given that \( f''(x) > 0 \) for all \( x \in \mathbb{R} \), we will follow these steps:

### Step 1: Differentiate \( g(x) \)
We start by finding the first derivative of \( g(x) \):
\[ g'(x) = \frac{d}{dx}[f(2 - x)] + \frac{d}{dx}[f(4 + x)] \]
Using the chain rule, we find:
\[ g'(x) = f'(2 - x) \cdot (-1) + f'(4 + x) \cdot (1) \]
This simplifies to:
\[ g'(x) = -f'(2 - x) + f'(4 + x) \]

### Step 2: Set the condition for \( g(x) \) to be increasing
For \( g(x) \) to be increasing, we need:
\[ g'(x) > 0 \]
This leads to:
\[ f'(4 + x) > f'(2 - x) \]

### Step 3: Analyze the implications of \( f''(x) > 0 \)
Since \( f''(x) > 0 \), it implies that \( f'(x) \) is an increasing function. Therefore, if \( a > b \), then \( f'(a) > f'(b) \).

### Step 4: Establish the inequality
We have \( 4 + x > 2 - x \). Rearranging gives:
\[ 4 + x > 2 - x \implies 2x > -2 \implies x > -1 \]

### Step 5: Conclusion
Thus, \( g(x) \) is increasing for:
\[ x > -1 \]
This means \( g(x) \) is increasing in the interval \( (-1, \infty) \).

### Final Answer
The function \( g(x) \) is increasing in the interval \( (-1, \infty) \).

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