Let `f: R rarr R` be a function defined as `f(x)=[(x+1)^2]^(1/3)+[(x-1)^2]^(1/3)+x/(2^x-1)+x/2+1`

Options

A. Even
B. Odd
C. Neither even nor odd
D. Both even and odd

Answer

To determine whether the function \( f(x) = \left( (x+1)^2 \right)^{1/3} + \left( (x-1)^2 \right)^{1/3} + \frac{x}{2^x - 1} + \frac{x}{2} + 1 \) is even, odd, or neither, we need to evaluate \( f(-x) \) and compare it with \( f(x) \) and \( -f(x) \).

### Step 1: Calculate \( f(-x) \)

Substituting \( -x \) into the function:

\[ f(-x) = \left( (-x + 1)^2 \right)^{1/3} + \left( (-x - 1)^2 \right)^{1/3} + \frac{-x}{2^{-x} - 1} + \frac{-x}{2} + 1 \]

### Step 2: Simplify \( f(-x) \)

Now we simplify each term:

1. **First term**:
\[ (-x + 1)^2 = (1 - x)^2 \implies \left( (1 - x)^2 \right)^{1/3} \]

2. **Second term**:
\[ (-x - 1)^2 = (-1 - x)^2 = (x + 1)^2 \implies \left( (x + 1)^2 \right)^{1/3} \]

3. **Third term**:
\[ \frac{-x}{2^{-x} - 1} = \frac{-x}{\frac{1}{2^x} - 1} = \frac{-x \cdot 2^x}{1 - 2^x} \]

4. **Fourth term**:
\[ \frac{-x}{2} \]

Putting it all together, we have:

\[ f(-x) = \left( (1 - x)^2 \right)^{1/3} + \left( (x + 1)^2 \right)^{1/3} + \frac{-x \cdot 2^x}{1 - 2^x} - \frac{x}{2} + 1 \]

### Step 3: Compare \( f(-x) \) with \( f(x) \)

Now we need to compare \( f(-x) \) with \( f(x) \):

1. **First term in \( f(x) \)**:
\[ \left( (x + 1)^2 \right)^{1/3} \]

2. **Second term in \( f(x) \)**:
\[ \left( (x - 1)^2 \right)^{1/3} \]

3. **Third term in \( f(x) \)**:
\[ \frac{x}{2^x - 1} \]

4. **Fourth term in \( f(x) \)**:
\[ \frac{x}{2} \]

5. **Constant term**:
\[ 1 \]

### Step 4: Check if \( f(-x) = f(x) \) or \( f(-x) = -f(x) \)

After simplifying \( f(-x) \), we can see that:

- The first term \( (1 - x)^2 \) does not equal \( (x + 1)^2 \).
- The second term \( (x + 1)^2 \) does not equal \( (x - 1)^2 \).
- The third term \( \frac{-x \cdot 2^x}{1 - 2^x} \) does not equal \( \frac{x}{2^x - 1} \).
- The fourth term \( -\frac{x}{2} \) does not equal \( \frac{x}{2} \).

Thus, \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \).

### Conclusion

Since \( f(-x) \) is neither equal to \( f(x) \) nor equal to \( -f(x) \), we conclude that the function \( f(x) \) is neither even nor odd.