Let `f (x)=|x-2|+|x - 3|+|x-4|` and `g(x) = f(x+1)`. Then 1. `g(x)` is an even function 2. `g(x)` is an odd function 3. `g(x)` is neither even nor odd 4. `g(x)` is periodic

Options

A. g(x) is an even function
B. g(x) is an odd function
C. g(x) is neither even nor odd
D. g(x) is periodic

Answer

To solve the problem, we need to analyze the function \( g(x) = f(x+1) \) where \( f(x) = |x-2| + |x-3| + |x-4| \).

### Step 1: Determine \( g(x) \)

We start by substituting \( x+1 \) into the function \( f(x) \):

\[ g(x) = f(x+1) = |(x+1) - 2| + |(x+1) - 3| + |(x+1) - 4| \]

This simplifies to:

\[ g(x) = |x - 1| + |x - 2| + |x - 3| \]

### Step 2: Determine \( g(-x) \)

Next, we calculate \( g(-x) \):

\[ g(-x) = f(-x + 1) = |(-x + 1) - 2| + |(-x + 1) - 3| + |(-x + 1) - 4| \]

This simplifies to:

\[ g(-x) = |-x - 1| + |-x - 2| + |-x - 3| = |-(x + 1)| + |-(x + 2)| + |-(x + 3)| \]

Which can be rewritten as:

\[ g(-x) = |x + 1| + |x + 2| + |x + 3| \]

### Step 3: Compare \( g(x) \) and \( g(-x) \)

Now we compare \( g(x) \) and \( g(-x) \):

- \( g(x) = |x - 1| + |x - 2| + |x - 3| \)
- \( g(-x) = |x + 1| + |x + 2| + |x + 3| \)

### Step 4: Check for Evenness

For \( g(x) \) to be an even function, we need \( g(x) = g(-x) \).

Clearly, \( g(x) \) and \( g(-x) \) are not equal for all \( x \) (for example, at \( x = 0 \), \( g(0) = 6 \) and \( g(-0) = 3 \)). Thus, \( g(x) \) is not even.

### Step 5: Check for Oddness

For \( g(x) \) to be an odd function, we need \( g(-x) = -g(x) \).

Calculating this for specific values will show that \( g(-x) \) does not equal \(-g(x)\) either. For example, at \( x = 0 \):

\[ g(0) = 6 \quad \text{and} \quad g(-0) = 3 \quad \text{and} \quad -g(0) = -6 \]

Thus, \( g(-x) \neq -g(x) \).

### Conclusion

Since \( g(x) \) is neither even nor odd, the correct answer is:

**3. \( g(x) \) is neither even nor odd.**