Question

The domain of the function $f(x)=\sqrt{x-1}$ is

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

Hint:

Logic Tip: Always check the restrictions of standard functions! For even roots (like square roots or 4th roots), the inside must be $\ge 0$. If you have a fraction, remember the denominator cannot equal $0$.

Answer 1

The Correct Option is C

Concept:
The domain of a function is the set of all possible input values (x-values) for which the function produces a valid real number output. For a square root function $f(x) = \sqrt{g(x)}$ to be defined in the set of real numbers, the expression inside the square root (the radicand) must be non-negative. That is, $g(x) \ge 0$.

Step 1: Set up the inequality.
In the given function $f(x) = \sqrt{x-1}$, the expression inside the square root is $x - 1$. Therefore, we must satisfy the condition: $$x - 1 \ge 0$$

Step 2: Solve for x.
Add 1 to both sides of the inequality to isolate $x$: $$x \ge 1$$ This means that for the function to work, $x$ can take any real value from 1 (inclusive) up to positive infinity.

Step 3: Express the result in interval notation.
The inequality $x \ge 1$ translates to the interval notation $[1, \infty)$. The square bracket "[" indicates that 1 is included, and the parenthesis ")" indicates that infinity is an open boundary.