The inverse of the function f(x) = frac{e^x - | vedclass

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(B) Let $y = f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} + 2$. Multiply the numerator and denominator of the fraction by $e^x$: $y = \frac{e^{2x} - 1}{e^

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$\log_e \left( \frac{x - 1}{3 - x} \right)^{1/2}$

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(B) Let $y = f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} + 2$. Multiply the numerator and denominator of the fraction by $e^x$: $y = \frac{e^{2x} - 1}{e^{2x} + 1} + 2$. Subtract $2$ from both sides: $y - 2 = \frac{e^{2x} - 1}{e^{2x} + 1}$. Let $u = e^{2x}$. Then $y - 2 = \frac{u - 1}{u + 1}$. $(y - 2)(u + 1) = u - 1$ $yu + y - 2u - 2 = u - 1$ $u(y - 2 - 1) = -1 - y + 2$ $u(y - 3) = 1 - y$ $u = \frac{1 - y}{y - 3} = \frac{y - 1}{3 - y}$. Since $u = e^{2x}$,we have $e^{2x} = \frac{y - 1}{3 - y}$. Taking the natural logarithm on both sides: $2x = \log_e \left( \frac{y - 1}{3 - y} \right)$ $x = \frac{1}{2} \log_e \left( \frac{y - 1}{3 - y} \right) = \log_e \left( \frac{y - 1}{3 - y} \right)^{1/2}$. Thus,the inverse function is $f^{-1}(x) = \log_e \left( \frac{x - 1}{3 - x} \right)^{1/2}$.

The inverse of the function $f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} + 2$ is given by

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