If a function `f:R rarr R` determined by equation `f(x)=-f(|x|)` us

Options

A. one - one
B. onto
C. many - one and into
D. non - differentiable

Answer

To solve the problem, we need to analyze the function defined by the equation \( f(x) = -f(|x|) \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function is defined such that for any real number \( x \), the value of the function at \( x \) is the negative of the value of the function at the absolute value of \( x \). This suggests that the function has a certain symmetry.

**Hint**: Consider the implications of the absolute value in the function definition.

2. **Evaluating the Function for Positive Values**:
Let's first consider when \( x > 0 \). In this case, \( |x| = x \). Therefore, we can substitute into the function:
\[ f(x) = -f(|x|) = -f(x) \]
This implies:
\[ f(x) + f(x) = 0 \quad \Rightarrow \quad 2f(x) = 0 \quad \Rightarrow \quad f(x) = 0 \]
So for all \( x > 0 \), \( f(x) = 0 \).

**Hint**: What happens to the function when you evaluate it at positive values?

3. **Evaluating the Function for Negative Values**:
Now consider when \( x < 0 \). In this case, \( |x| = -x \). Thus:
\[ f(x) = -f(|x|) = -f(-x) \]
Since we already determined that \( f(-x) = 0 \) for \( -x > 0 \) (which means \( x < 0 \)), we have:
\[ f(x) = -0 = 0 \]
So for all \( x < 0 \), \( f(x) = 0 \).

**Hint**: How does the function behave for negative values based on the previous result?

4. **Evaluating the Function at Zero**:
Now, let's evaluate the function at \( x = 0 \):
\[ f(0) = -f(|0|) = -f(0) \]
This leads to:
\[ f(0) + f(0) = 0 \quad \Rightarrow \quad 2f(0) = 0 \quad \Rightarrow \quad f(0) = 0 \]

**Hint**: What can you conclude about the function at zero?

5. **Conclusion**:
From the evaluations above, we find that:
\[ f(x) = 0 \quad \text{for all } x \in \mathbb{R} \]
Thus, the function is the zero function.

**Hint**: What does this imply about the nature of the function in terms of injectivity and surjectivity?

### Final Result:
The function \( f(x) \) is identically zero for all \( x \in \mathbb{R} \).