Q.(i) [2] Draw the graph of the function, $f : \mathbb{R} - \{0\} \to \mathbb{R}$ defined by $f(x) = \dfrac{1}{x}$.

What is a Rational Function Domain?

Imagine you're baking a cake and the recipe says "add flour until the mixture is smooth." If you add too much flour, the mixture becomes a dry lump — it stops being a proper batter. A rational function is like that mixture: it's a fraction made of two polynomials, and it only "works" when the denominator isn't zero.

A rational function looks like this:

where $P(x)$ and $Q(x)$ are polynomials, and $Q(x) \neq 0$.

The domain of a rational function is simply the set of all real numbers $x$ for which the function is defined — meaning, all $x$ except those that make the denominator zero.

The Intuition First

Think of division in everyday life. You can divide 10 apples among 5 people — that's fine. You can divide 10 apples among 2 people — also fine. But can you divide 10 apples among 0 people? That doesn't make sense. You can't split something among nobody.

In the same way, a rational function is a division. The denominator tells you "how many groups" you're splitting into. If the denominator is zero, the division is impossible — the function has no value there.

So the domain is: all real numbers, except the ones that make the bottom zero.

The Precise Statement

In plain words: find every $x$ that makes $Q(x) = 0$, and remove those from the set of all real numbers.

How to Find the Domain — Step by Step

Step 1: Write down the denominator $Q(x)$.

Step 2: Set $Q(x) = 0$ and solve for $x$.

Step 3: The domain is all real numbers except those solutions.

Examples

Example 1: $f(x) = \frac{1}{x-3}$

Denominator: $x - 3 = 0 \implies x = 3$

Domain: all real numbers except $3$. In interval notation: $(-\infty, 3) \cup (3, \infty)$

Example 2: $f(x) = \frac{x^2 + 1}{x^2 - 4}$

Denominator: $x^2 - 4 = 0 \implies (x-2)(x+2) = 0 \implies x = 2$ or $x = -2$

Domain: all real numbers except $2$ and $-2$. In interval notation: $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$

Example 3: $f(x) = \frac{2x+5}{x^2 + 1}$

Denominator: $x^2 + 1 = 0 \implies x^2 = -1$ — no real solution.

Domain: all real numbers, i.e., $(-\infty, \infty)$ …