If f(x)=`root3(1-x^2)+root3(1+x^2)`, then f(x) is :

Options

A. an even function
B. an odd function
C. constant function
D. none of these

Answer

To determine the nature of the function \( f(x) = \sqrt[3]{1 - x^2} + \sqrt[3]{1 + x^2} \), we will check if it is an even function, an odd function, or neither.

### Step-by-Step Solution:

1. **Define the Function**:
We have the function:
\[ f(x) = \sqrt[3]{1 - x^2} + \sqrt[3]{1 + x^2} \]

2. **Find \( f(-x) \)**:
To check if the function is even or odd, we need to calculate \( f(-x) \):
\[ f(-x) = \sqrt[3]{1 - (-x)^2} + \sqrt[3]{1 + (-x)^2} \]
Since \((-x)^2 = x^2\), we can simplify this to:
\[ f(-x) = \sqrt[3]{1 - x^2} + \sqrt[3]{1 + x^2} \]

3. **Compare \( f(-x) \) with \( f(x) \)**:
Now, we compare \( f(-x) \) with \( f(x) \):
\[ f(-x) = \sqrt[3]{1 - x^2} + \sqrt[3]{1 + x^2} = f(x) \]

4. **Conclusion**:
Since \( f(-x) = f(x) \), we conclude that \( f(x) \) is an even function.

### Final Answer:
Thus, the function \( f(x) \) is an **even function**.