Questions: Find the inverse function for f(x) : f(x) = (e^x - 3e^(-x))/2

Solution Steps

To find the inverse function of \( f(x) = \frac{e^x - 3e^{-x}}{2} \), we need to follow these steps:

1. Set \( y = \frac{e^x - 3e^{-x}}{2} \).
2. Solve for \( x \) in terms of \( y \).
3. Swap \( x \) and \( y \) to express the inverse function \( f^{-1}(x) \).

We start with the function defined as: \[ y = f(x) = \frac{e^x - 3e^{-x}}{2} \] To find the inverse function, we need to express \( x \) in terms of \( y \).

Rearranging the equation gives us: \[ 2y = e^x - 3e^{-x} \] Multiplying through by \( e^x \) to eliminate the negative exponent results in: \[ 2ye^x = e^{2x} - 3 \] This can be rewritten as: \[ e^{2x} - 2ye^x - 3 = 0 \]

This is a quadratic equation in terms of \( e^x \). We can apply the quadratic formula: \[ e^x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -2y \), and \( c = -3 \). Thus, we have: \[ e^x = \frac{2y \pm \sqrt{(2y)^2 + 12}}{2} \] This simplifies to: \[ e^x = y \pm \sqrt{y^2 + 3} \]

Taking the natural logarithm of both sides gives us two potential solutions for \( x \): \[ x = \log(y - \sqrt{y^2 + 3}) \quad \text{and} \quad x = \log(y + \sqrt{y^2 + 3}) \]

Final Answer