Let $g(x) = \frac{e^x - e^{-x}}{2}$ and $g(f(x)) = x$, then evaluate $f\left(\frac{e^{22} - 1}{2e^{11}}\right)$.
Step-by-Step Solution
Key Concept: Recognize that g(x) = (e^x - e^(-x))/2 is the hyperbolic sine function sinh(x), and finding f requires computing the inverse function g^(-1)(x) = ln(x + √(x² + 1)), then evaluating this at the given argument.
Given $y = g(x) = \frac{e^x - e^{-x}}{2}$, we solve for the inverse by setting $e^{2x} - 1 = 2ye^x$. Substituting $t = e^x$ yields $t^2 - 2yt - 1 = 0$, giving $t = y \pm \sqrt{y^2+1}$. Since $e^x > 0$, we take $e^x = y + \sqrt{y^2+1}$, so $g^{-1}(x) = f(x) = \ln(x + \sqrt{x^2+1})$. Finally, $f\left(\frac{e^{22}-1}{2e^{11}}\right) = \ln\left[\left(\frac{e^{22}-1}{2e^{11}}\right)^2 + 1\right]^{1/2} = \ln(e^{11}) = 11$.
Correct Answer: 1