The domain of the function f(x) = frac{1}{1 + | vedclass

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(A) Given the domain is $-1 \le x \le 1$.Since $f(x) = \frac{1}{1 + e^x}$ is a strictly decreasing function,the range is $[f(1), f(-1)]$.Calculating t

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$\left( \frac{1}{1+e}, \frac{1}{1+e^{-1}} \right)$

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(A) Given the domain is $-1 \le x \le 1$.Since $f(x) = \frac{1}{1 + e^x}$ is a strictly decreasing function,the range is $[f(1), f(-1)]$.Calculating the values:$f(1) = \frac{1}{1 + e^1} = \frac{1}{1 + e}$$f(-1) = \frac{1}{1 + e^{-1}} = \frac{1}{1 + \frac{1}{e}} = \frac{e}{e + 1}$Thus,the range is $\left[ \frac{1}{1+e}, \frac{e}{1+e} \right]$.

The domain of the function $f(x) = \frac{1}{1 + e^x}$ is $[-1, 1]$. Find the range of the function.

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