Range of the `f(x) = (e^(x) - 1)/(e^(x) + 1)`

Options

A. `(0, oo)`
B. `(-oo, 0)`
C. `(1, oo)`
D. `(-1, 1)`

Answer

To find the range of the function \( f(x) = \frac{e^x - 1}{e^x + 1} \), we can follow these steps:

### Step 1: Set the function equal to \( y \)
We start by letting \( f(x) = y \):
\[ y = \frac{e^x - 1}{e^x + 1} \]

### Step 2: Cross-multiply to eliminate the fraction
To eliminate the fraction, we cross-multiply:
\[ y(e^x + 1) = e^x - 1 \]
This simplifies to:
\[ ye^x + y = e^x - 1 \]

### Step 3: Rearrange the equation
Rearranging the equation gives:
\[ ye^x - e^x = -1 - y \]
Factoring out \( e^x \) from the left side, we have:
\[ e^x(y - 1) = -1 - y \]

### Step 4: Solve for \( e^x \)
Now we can solve for \( e^x \):
\[ e^x = \frac{-1 - y}{y - 1} \]

### Step 5: Take the logarithm of both sides
Taking the natural logarithm of both sides gives:
\[ x = \ln\left(\frac{-1 - y}{y - 1}\right) \]

### Step 6: Determine the conditions for \( e^x \)
Since \( e^x \) is always positive, we need:
\[ \frac{-1 - y}{y - 1} > 0 \]

### Step 7: Analyze the inequality
To analyze the inequality \( \frac{-1 - y}{y - 1} > 0 \), we find the critical points by setting the numerator and denominator to zero:
1. \( -1 - y = 0 \) gives \( y = -1 \)
2. \( y - 1 = 0 \) gives \( y = 1 \)

### Step 8: Test intervals around the critical points
We test the intervals defined by the points \( y = -1 \) and \( y = 1 \):
- For \( y < -1 \): both numerator and denominator are negative, so the fraction is positive.
- For \( -1 < y < 1 \): the numerator is positive and the denominator is negative, so the fraction is negative.
- For \( y > 1 \): both numerator and denominator are negative, so the fraction is positive.

### Step 9: Conclusion about the range
From the analysis, the function is defined for the interval \( -1 < y < 1 \). Therefore, the range of \( f(x) \) is:
\[ \text{Range of } f(x) = (-1, 1) \]

### Final Answer
Thus, the range of the function \( f(x) = \frac{e^x - 1}{e^x + 1} \) is:
\[ (-1, 1) \]

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