For the function f(x) = (1 + frac{1}{x})^x,wh | vedclass

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(D) Let $f(x) = (1 + \frac{1}{x})^x$. The domain of $f(x)$ is $(-\infty, -1) \cup (0, \infty)$.Taking the natural logarithm,$\ln(f(x)) = x \ln(1 + \fr

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has neither a maxima nor a minima

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(D) Let $f(x) = (1 + \frac{1}{x})^x$. The domain of $f(x)$ is $(-\infty, -1) \cup (0, \infty)$.Taking the natural logarithm,$\ln(f(x)) = x \ln(1 + \frac{1}{x})$.Differentiating with respect to $x$,$\frac{f'(x)}{f(x)} = \ln(1 + \frac{1}{x}) + x \cdot \frac{1}{1 + \frac{1}{x}} \cdot (-\frac{1}{x^2}) = \ln(1 + \frac{1}{x}) - \frac{1}{x+1}$.Let $g(x) = \ln(1 + \frac{1}{x}) - \frac{1}{x+1}$.For $x > 0$,let $t = \frac{1}{x}$,then $t \in (0, \infty)$. $g(t) = \ln(1+t) - \frac{t}{1+t}$.$g'(t) = \frac{1}{1+t} - \frac{(1+t) - t}{(1+t)^2} = \frac{1}{1+t} - \frac{1}{(1+t)^2} = \frac{t}{(1+t)^2} > 0$ for $t > 0$.Since $g(0) = 0$ and $g(t)$ is strictly increasing,$g(x) > 0$ for all $x > 0$.Thus $f'(x) > 0$ for all $x > 0$,meaning $f(x)$ is strictly increasing on $(0, \infty)$.For $x  1$. $f(x) = (1 - \frac{1}{u})^{-u} = (\frac{u-1}{u})^{-u} = (\frac{u}{u-1})^u$.Analysis shows $f'(x)$ does not vanish in the domain,so there are no local extrema.

For the function $f(x) = (1 + \frac{1}{x})^x$,which of the following is true?

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