The largest possible set of real numbers whic | vedclass

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Question

(D) For the function $f(x) = \sqrt{1 - \frac{1}{x}}$ to be defined,the expression inside the square root must be non-negative:$1 - \frac{1}{x} \ge 0$$

Vedclass · Accepted Answer

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

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(D) For the function $f(x) = \sqrt{1 - \frac{1}{x}}$ to be defined,the expression inside the square root must be non-negative:$1 - \frac{1}{x} \ge 0$$\frac{x - 1}{x} \ge 0$Using the wavy curve method (sign scheme) for the critical points $x = 0$ and $x = 1$:The expression is positive in the intervals $(-\infty, 0)$ and $[1, \infty)$.Note that $x = 0$ is excluded because the denominator cannot be zero.Thus,the domain is $(-\infty, 0) \cup [1, \infty)$.

The largest possible set of real numbers which can be the domain of $f(x) = \sqrt{1 - \frac{1}{x}}$ is

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