If `f(x)=root(3)((1-x^(2)))+root(3)((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 whether it is an even function, odd function, constant function, or none of these.

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

### Step 2: Check for evenness
To check if \( f(x) \) is an even function, we need to evaluate \( f(-x) \) and see if it equals \( f(x) \).

Calculating \( f(-x) \):
\[ f(-x) = \sqrt[3]{1 - (-x)^2} + \sqrt[3]{1 + (-x)^2} \]
Since \( (-x)^2 = x^2 \), we can simplify:
\[ f(-x) = \sqrt[3]{1 - x^2} + \sqrt[3]{1 + x^2} \]
This simplifies to:
\[ f(-x) = \sqrt[3]{1 - x^2} + \sqrt[3]{1 + x^2} = f(x) \]

### Step 3: Conclusion
Since \( f(-x) = f(x) \), we conclude that \( f(x) \) is an even function.

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

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