Question

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

• A. (0, 1) $\cup$ (0, $\infty$)
• B. (-1, 0) $\cup$ (1, $\infty$)
• C. (-$\infty$, -1) $\cup$ (0, $\infty$)
• D. (-$\infty$, 0) $\cup$ (1, $\infty$)

Answer 1

Answer: (D)

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

Answer 2

To find the domain of the function $f(x) = \sqrt{1-\frac{1}{x}}$, we need to ensure that the expression under the square root is non-negative and the denominator of any fraction is not zero.

We require $1-\frac{1}{x} \ge 0$, which simplifies to $\frac{x-1}{x} \ge 0$.

The critical points are $x=0$ and $x=1$. We analyze the intervals determined by these points:

• Interval $(-\infty, 0)$: Choose $x = -1$. Then $\frac{-1-1}{-1} = 2 \ge 0$. This interval is part of the domain.

• Interval $(0, 1)$: Choose $x = 0.5$. Then $\frac{0.5-1}{0.5} = -1 < 0$. This interval is not part of the domain.

• Interval $(1, \infty)$: Choose $x = 2$. Then $\frac{2-1}{2} = \frac{1}{2} \ge 0$. This interval is part of the domain.

We also need to consider the critical points themselves:

• $x=0$ is excluded because it makes the denominator zero in the original function.

• $x=1$ is included because $f(1) = \sqrt{1-\frac{1}{1}} = \sqrt{0} = 0$, which is a real number.

Therefore, the domain of $f(x)$ is $(-\infty, 0) \cup [1, \infty)$.

Among the given options, $(-\infty, 0) \cup (1, \infty)$ is the closest, although it incorrectly excludes $x=1$.