If the function of `f:R->A` is given by `f(x)=(e^x-e^(-|x|))/(e^x+e^(|x|))` is surjection, find A

Answer

To determine the set \( A \) such that the function \( f: \mathbb{R} \to A \) defined by

\[ f(x) = \frac{e^x - e^{-|x|}}{e^x + e^{|x|}} \]

is a surjection, we need to analyze the behavior of the function for both cases of \( x \) (i.e., \( x \geq 0 \) and \( x < 0 \)).

### Step 1: Analyze the function for \( x \geq 0 \)

For \( x \geq 0 \), we have \( |x| = x \). Thus, the function simplifies to:

\[ f(x) = \frac{e^x - e^{-x}}{e^x + e^x} = \frac{e^x - e^{-x}}{2e^x} \]

This can be further simplified:

\[ f(x) = \frac{1 - e^{-2x}}{2} \]

### Step 2: Determine the range for \( x \geq 0 \)

As \( x \) approaches \( 0 \):

\[ f(0) = \frac{1 - e^{0}}{2} = \frac{1 - 1}{2} = 0 \]

As \( x \) approaches \( \infty \):

\[ f(x) \to \frac{1 - 0}{2} = \frac{1}{2} \]

Thus, for \( x \geq 0 \), the range of \( f(x) \) is \( [0, \frac{1}{2}) \).

### Step 3: Analyze the function for \( x < 0 \)

For \( x < 0 \), we have \( |x| = -x \). Thus, the function becomes:

\[ f(x) = \frac{e^x - e^{x}}{e^x + e^{-x}} = \frac{0}{e^x + e^{-x}} = 0 \]

### Step 4: Combine the results

From the analysis, we find that:

- For \( x \geq 0 \), \( f(x) \) takes values in \( [0, \frac{1}{2}) \).
- For \( x < 0 \), \( f(x) = 0 \).

### Step 5: Determine the codomain \( A \)

Since \( f(x) \) can take any value from \( 0 \) to just below \( \frac{1}{2} \) (not including \( \frac{1}{2} \)), the codomain \( A \) must be:

\[ A = [0, \frac{1}{2}) \]

### Conclusion

Thus, the set \( A \) for which the function \( f \) is a surjection is:

\[ \boxed{[0, \frac{1}{2})} \]