Let `f^(prime)(x)>0 AA x in R and g(x)=f(2-x)+f(4+x).` Then g(x) is increasing in (i) `(-infty,-1)` (ii) `(-infty,0)` (iii) `(-1,infty)` (iv) none of these

Options

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

Answer

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

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

### Step 2: Set the derivative greater than zero
For \( g(x) \) to be increasing, we need:
\[ g'(x) > 0 \]
This implies:
\[ f'(4 + x) > f'(2 - x) \]

### Step 3: Analyze the condition
Since we know that \( f'(x) > 0 \) for all \( x \in \mathbb{R} \), it means that \( f(x) \) is an increasing function. Therefore, if \( f'(4 + x) > f'(2 - x) \), it indicates that the argument \( 4 + x \) is greater than the argument \( 2 - x \).

### Step 4: Set up the inequality
We set up the inequality:
\[ 4 + x > 2 - x \]
Solving this gives:
\[ 4 + x + x > 2 \]
\[ 2x > 2 - 4 \]
\[ 2x > -2 \]
\[ x > -1 \]

### Conclusion
Thus, \( g(x) \) is increasing for \( x > -1 \). Therefore, the correct interval where \( g(x) \) is increasing is \( (-1, \infty) \).

### Final Answer
The correct option is (iii) \( (-1, \infty) \).

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