The largest possible set of real numbers whic | vedclass

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(D) For the function $f(x) = \sqrt{1 - \frac{1}{x}}$ to be defined,the expression under 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 under the square root must be non-negative:$1 - \frac{1}{x} \ge 0$$\frac{x - 1}{x} \ge 0$To solve this inequality,we find the critical points where the numerator and denominator are zero,which are $x = 1$ and $x = 0$.Using the sign scheme (Wavy Curve Method):For $x > 1$,$\frac{x-1}{x} > 0$ (positive).For $0 For $x  0$ (positive).Since we need the expression to be $\ge 0$,the solution is $x \in (-\infty, 0) \cup [1, \infty)$.Note: $x = 0$ is excluded because the denominator cannot be zero.

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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