Find whether the following functions are one-one or many-one `f(x)=sqrt(1-e^(1/x-1))`

Explanation

To determine whether the function \( f(x) = \sqrt{1 - e^{(1/x - 1)}} \) is one-one or many-one, we will follow these steps:

### Step 1: Analyze the Function
The function is given as:
\[ f(x) = \sqrt{1 - e^{(1/x - 1)}} \]
We need to ensure that the expression inside the square root is non-negative:
\[ 1 - e^{(1/x - 1)} \geq 0 \]
This implies:
\[ e^{(1/x - 1)} \leq 1 \]
Taking the natural logarithm on both sides, we have:
\[ \frac{1}{x} - 1 \leq 0 \implies \frac{1}{x} \leq 1 \implies x \geq 1 \]

### Step 2: Determine the Domain
From the inequality \( x \geq 1 \), the domain of \( f(x) \) is:
\[ x \in [1, \infty) \]

### Step 3: Differentiate the Function
To check if the function is one-one, we can differentiate \( f(x) \):
\[ f(x) = \sqrt{1 - e^{(1/x - 1)}} \]
Let’s rewrite it for easier differentiation:
\[ f(x) = (1 - e^{(1/x - 1)})^{1/2} \]
Using the chain rule:
\[ f'(x) = \frac{1}{2}(1 - e^{(1/x - 1)})^{-1/2} \cdot \frac{d}{dx}(1 - e^{(1/x - 1)}) \]
Now, we need to differentiate \( 1 - e^{(1/x - 1)} \):
\[ \frac{d}{dx}(1 - e^{(1/x - 1)}) = -e^{(1/x - 1)} \cdot \frac{d}{dx}(1/x - 1) \]
Calculating \( \frac{d}{dx}(1/x) \):
\[ \frac{d}{dx}(1/x) = -\frac{1}{x^2} \]
Thus,
\[ \frac{d}{dx}(1 - e^{(1/x - 1)}) = -e^{(1/x - 1)} \cdot \left(-\frac{1}{x^2}\right) = \frac{e^{(1/x - 1)}}{x^2} \]
Now substituting back into \( f'(x) \):
\[ f'(x) = \frac{1}{2}(1 - e^{(1/x - 1)})^{-1/2} \cdot \frac{e^{(1/x - 1)}}{x^2} \]

### Step 4: Analyze the Sign of the Derivative
Since \( e^{(1/x - 1)} > 0 \) for \( x \geq 1 \) and \( 1 - e^{(1/x - 1)} \) is non-negative in the domain, we can conclude:
- \( f'(x) > 0 \) for \( x \in [1, \infty) \)

### Step 5: Conclusion
Since \( f'(x) > 0 \) in the domain \( [1, \infty) \), the function \( f(x) \) is strictly increasing. Therefore, \( f(x) \) is a one-one function.

### Final Answer
The function \( f(x) = \sqrt{1 - e^{(1/x - 1)}} \) is a one-one function.

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