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(C) For the function $f(x) = \left(1 + \frac{1}{x}\right)^x$ to be defined as a real-valued function,the base must be positive for all real $x$ in the

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) For the function $f(x) = \left(1 + \frac{1}{x}\right)^x$ to be defined as a real-valued function,the base must be positive for all real $x$ in the domain.We require $1 + \frac{1}{x} > 0$.This simplifies to $\frac{x + 1}{x} > 0$.Using the sign scheme (Wavy curve method) for the critical points $x = -1$ and $x = 0$:For $x  0$ (e.g.,$x = -2 \implies \frac{-1}{-2} = 0.5 > 0$).For $-1 For $x > 0$,$\frac{x+1}{x} > 0$ (e.g.,$x = 1 \implies \frac{2}{1} = 2 > 0$).Thus,the domain is $x \in (-\infty, -1) \cup (0, \infty)$.

The set of all real values of $x$ for which the real-valued function $f(x) = \left(1 + \frac{1}{x}\right)^x$ is defined,is

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