Q.Find the domain and the range of the real function $f$ defined by $f(x) = \sqrt{x - 1}$.

Concept: Rational Function Domain — but here it's a square root function, so the domain is all $x$ for which the radicand is non-negative.

Step 1 – Domain condition:

We need $x - 1 \ge 0$, i.e. $x \ge 1$.

Thus domain $= [1, \infty)$.

Step 2 – Range reasoning: …

Concept and Intuition

When we talk about a real function, we mean that both the input $x$ and the output $f(x)$ must be real numbers. The square root of a negative number is not real — it belongs to the complex numbers. So the first question is: for which real $x$ does $\sqrt{x-1}$ give a real result?

That’s the domain — the set of all permissible inputs.

Once we know what $x$ values are allowed, we ask: what values does $f(x)$ actually take? That’s the range — the set of all possible outputs.

The key idea: the square root function $\sqrt{t}$ (for real $t$) is only defined when $t \geq 0$, and its output is always $0$ or positive. So we translate the condition $x-1 \geq 0$ into the domain, and then think about what numbers the square root can produce.

Step-by-step solution

1. Find the domain For $f(x) = \sqrt{x-1}$ to be real, the expression inside the square root must be non-negative:

Solving this gives:

So the domain is all real numbers from $1$ to infinity. In interval notation:

2. Find the range We now ask: as $x$ takes every value in $[1, \infty)$, what values does $f(x) = \sqrt{x-1}$ produce?
   • When $x = 1$, we get $f(1) = \sqrt{0} = 0$.
   • As $x$ increases, $x-1$ increases, and the square root of a larger number is larger. …