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

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Question

(C) For the function $f(x) = (1 + \frac{1}{x})^x$ to be defined,the base must be positive,i.e.,$1 + \frac{1}{x} > 0$. Solving the inequality: $\frac{x

Vedclass · Accepted Answer

$(-\infty, -1) \cup (0, \infty)$

Vedclass · Answer

(C) For the function $f(x) = (1 + \frac{1}{x})^x$ to be defined,the base must be positive,i.e.,$1 + \frac{1}{x} > 0$. Solving the inequality: $\frac{x+1}{x} > 0$. Using the sign scheme method,we identify the critical points at $x = -1$ and $x = 0$. The expression $\frac{x+1}{x}$ is positive in the intervals $(-\infty, -1)$ and $(0, \infty)$. Thus,the domain of the function is $(-\infty, -1) \cup (0, \infty)$.

For the function $f(x) = (1 + \frac{1}{x})^x$,the domain of $f(x)$ is:

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