Given the function f(x) = left( frac{e^{2x} - | vedclass

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(A) Given $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$.To check if the function is increasing or decreasing,we find its derivative $f'(x)$ using the quotien

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Increasing

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(A) Given $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$.To check if the function is increasing or decreasing,we find its derivative $f'(x)$ using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$.Let $u = e^{2x} - 1$ and $v = e^{2x} + 1$. Then $u' = 2e^{2x}$ and $v' = 2e^{2x}$.$f'(x) = \frac{(2e^{2x})(e^{2x} + 1) - (e^{2x} - 1)(2e^{2x})}{(e^{2x} + 1)^2}$$f'(x) = \frac{2e^{4x} + 2e^{2x} - (2e^{4x} - 2e^{2x})}{(e^{2x} + 1)^2}$$f'(x) = \frac{4e^{2x}}{(e^{2x} + 1)^2}$.Since $e^{2x} > 0$ for all $x \in \mathbb{R}$,$f'(x) > 0$ for all $x \in \mathbb{R}$.Therefore,$f(x)$ is an increasing function.

Given the function $f(x) = \left( \frac{e^{2x} - 1}{e^{2x} + 1} \right)$,the function is:

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