`R rarr R` be defined by `f(x)=((e^(2x)-e^(-2x)))/2`. is `f(x)` invertible. If yes then find `f^(-1)(x)`

Explanation

To determine if the function \( f(x) = \frac{e^{2x} - e^{-2x}}{2} \) is invertible and to find its inverse if it is, we will follow these steps:

### Step 1: Check if the function is one-to-one (1-1)

A function is one-to-one if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).

Assume \( f(x_1) = f(x_2) \):
\[ \frac{e^{2x_1} - e^{-2x_1}}{2} = \frac{e^{2x_2} - e^{-2x_2}}{2} \]
This simplifies to:
\[ e^{2x_1} - e^{-2x_1} = e^{2x_2} - e^{-2x_2} \]
Rearranging gives:
\[ e^{2x_1} - e^{2x_2} = e^{-2x_1} - e^{-2x_2} \]
Factoring both sides:
\[ (e^{2x_1} - e^{2x_2}) = (e^{-2x_1} - e^{-2x_2}) \]
Let \( a = e^{2x_1} \) and \( b = e^{2x_2} \). Then:
\[ a - b = \frac{1}{a} - \frac{1}{b} \]
Cross-multiplying gives:
\[ (a - b)ab = b - a \]
This implies:
\[ (a - b)(ab + 1) = 0 \]
Thus, either \( a = b \) or \( ab + 1 = 0 \). Since \( a \) and \( b \) are both positive (as they are exponentials), we must have \( a = b \), which means \( e^{2x_1} = e^{2x_2} \) and therefore \( 2x_1 = 2x_2 \) leading to \( x_1 = x_2 \). Hence, \( f \) is one-to-one.

### Step 2: Check if the function is onto (surjective)

To check if \( f \) is onto, we need to show that for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \).

Setting \( y = f(x) \):
\[ y = \frac{e^{2x} - e^{-2x}}{2} \]
This can be rewritten as:
\[ 2y = e^{2x} - e^{-2x} \]
Multiplying through by \( e^{2x} \) gives:
\[ 2ye^{2x} = e^{4x} - 1 \]
Rearranging leads to the quadratic equation in \( e^{2x} \):
\[ e^{4x} - 2ye^{2x} - 1 = 0 \]
Let \( t = e^{2x} \):
\[ t^2 - 2yt - 1 = 0 \]
Using the quadratic formula:
\[ t = \frac{2y \pm \sqrt{(2y)^2 + 4}}{2} = y \pm \sqrt{y^2 + 1} \]
Since \( e^{2x} > 0 \), we take the positive root:
\[ e^{2x} = y + \sqrt{y^2 + 1} \]
This shows that for every \( y \), there exists an \( x \) such that \( f(x) = y \). Thus, \( f \) is onto.

### Conclusion

Since \( f \) is both one-to-one and onto, it is invertible.

### Step 3: Find the inverse function \( f^{-1}(x) \)

From our previous steps, we have:
\[ e^{2x} = x + \sqrt{x^2 + 1} \]
Taking the natural logarithm:
\[ 2x = \ln(x + \sqrt{x^2 + 1}) \]
Thus:
\[ x = \frac{1}{2} \ln(x + \sqrt{x^2 + 1}) \]
Therefore, the inverse function is:
\[ f^{-1}(x) = \frac{1}{2} \ln(x + \sqrt{x^2 + 1}) \]