The function f : R rightarrow (-1, 1) defined | vedclass

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(B) Given $f(x) = \frac{e^x - 1}{e^x + 1}$.We can rewrite the function as $f(x) = \frac{e^x + 1 - 2}{e^x + 1} = 1 - \frac{2}{e^x + 1}$.To check for on

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one-one onto

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(B) Given $f(x) = \frac{e^x - 1}{e^x + 1}$.We can rewrite the function as $f(x) = \frac{e^x + 1 - 2}{e^x + 1} = 1 - \frac{2}{e^x + 1}$.To check for one-one,we find the derivative: $f'(x) = \frac{d}{dx} (1 - 2(e^x + 1)^{-1}) = 0 - 2(-1)(e^x + 1)^{-2} \cdot e^x = \frac{2e^x}{(e^x + 1)^2}$.Since $e^x > 0$ for all $x \in R$,$f'(x) > 0$ for all $x \in R$. Thus,$f(x)$ is strictly increasing and therefore one-one.To check for onto,we look at the range. As $x \rightarrow \infty$,$f(x) \rightarrow 1$. As $x \rightarrow -\infty$,$f(x) \rightarrow -1$. Since the function is continuous and strictly increasing,the range is $(-1, 1)$,which is equal to the codomain.Therefore,the function is one-one and onto.

The function $f : R \rightarrow (-1, 1)$ defined by $f(x) = \frac{e^x - 1}{e^x + 1}$ is:

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