Let g: (-infty, infty) to (-frac{pi}{2}, frac | vedclass

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(C) For the function $g(x) = 2 	an^{-1}(e^x) - \frac{\pi}{2}$,we find its derivative: $g'(x) = 2 \cdot \frac{1}{1 + (e^x)^2} \cdot e^x = \frac{2e^x}{

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An odd function and strictly increasing on $(-\infty, \infty)$.

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(C) For the function $g(x) = 2 	an^{-1}(e^x) - \frac{\pi}{2}$,we find its derivative: $g'(x) = 2 \cdot \frac{1}{1 + (e^x)^2} \cdot e^x = \frac{2e^x}{1 + e^{2x}}$. Since $e^x > 0$ for all $x \in (-\infty, \infty)$,it follows that $g'(x) > 0$. Thus,$g(x)$ is a strictly increasing function on $(-\infty, \infty)$. Now,check for parity: $g(-x) = 2 	an^{-1}(e^{-x}) - \frac{\pi}{2} = 2 	an^{-1}(\frac{1}{e^x}) - \frac{\pi}{2}$. Using the identity $	an^{-1}(\frac{1}{u}) = \frac{\pi}{2} - 	an^{-1}(u)$ for $u > 0$: $g(-x) = 2(\frac{\pi}{2} - 	an^{-1}(e^x)) - \frac{\pi}{2} = \pi - 2 	an^{-1}(e^x) - \frac{\pi}{2} = \frac{\pi}{2} - 2 	an^{-1}(e^x) = -g(x)$. Since $g(-x) = -g(x)$,the function $g(x)$ is an odd function.

Let $g: (-\infty, \infty) \to (-\frac{\pi}{2}, \frac{\pi}{2})$ be defined by $g(x) = 2 \tan^{-1}(e^x) - \frac{\pi}{2}$. Then $g(x)$ is...

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