The function f(x) = frac{e^{2x} - 1}{e^{2x} + | vedclass

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

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Increasing

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

The function $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$ is a ........ function.

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