Function defined by `f(x) =(e^(x^(2))-e^(-x^(2)))/(e^(x^(2))+e^(-x^(2)))` is injective in `[alpha -2 ,oo)` the least value of `alpha` is ____

Answer

To determine the least value of \( \alpha \) such that the function

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

is injective on the interval \([ \alpha - 2, \infty )\), we will analyze the function step by step.

### Step 1: Understanding the Function
The function \( f(x) \) can be rewritten using the hyperbolic tangent function:
\[ f(x) = \tanh(x^2) \]
This is because:
\[ \tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} \]
Thus, we have:
\[ f(x) = \tanh(x^2) \]

### Step 2: Analyzing Injectivity
A function is injective if it is either strictly increasing or strictly decreasing. To check this, we will find the derivative of \( f(x) \).

### Step 3: Finding the Derivative
Using the chain rule, we find the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx} \tanh(x^2) = 2x \cdot \text{sech}^2(x^2) \]
Here, \( \text{sech}^2(x^2) \) is always positive for all \( x \). Therefore, \( f'(x) \) is positive for \( x > 0 \) and negative for \( x < 0 \).

### Step 4: Determining the Domain for Injectivity
Since \( f(x) \) is strictly increasing for \( x > 0 \), it will be injective in any interval that starts from a point greater than or equal to 0.

### Step 5: Setting the Domain
We need the interval \([ \alpha - 2, \infty )\) to start from 0. Thus, we set:
\[ \alpha - 2 = 0 \implies \alpha = 2 \]

### Conclusion
The least value of \( \alpha \) such that the function \( f(x) \) is injective in the interval \([ \alpha - 2, \infty )\) is:
\[ \boxed{2} \]