Q.Find the domain and range of the function $f(x) = \dfrac{1}{x-3}$.

This is a rational function, so we must exclude the value of $x$ that makes the denominator zero, and check which values the output can actually take.

The denominator $x-3$ is zero when $x=3$, so $x=3$ must be excluded from the domain: Domain $= \mathbb{R} - \{3\}$. Setting $y=\dfrac{1}{x-3}$ and solving for $x$ gives $x = 3+\dfrac1y$, which is defined for every $y$ except $y=0$; so the output $y$ can never equal $0$.

Step 1 — Domain. $f(x) = \dfrac{1}{x-3}$ is undefined exactly when its denominator is zero: $x-3=0 \Rightarrow x=3$. So $f$ is defined for every real $x$ except $3$: Domain $=\mathbb{R}-\{3\}$.

Step 2 — Range. Let $y=f(x)=\dfrac{1}{x-3}$. Solve for $x$ in terms of $y$: $y(x-3)=1 \Rightarrow x = 3+\dfrac1y$. This expression for $x$ is defined for every real $y$ except $y=0$ (division by $y$). So every real number except $0$ is attained by $f$, and $0$ itself is never attained (there is no $x$ with $\frac{1}{x-3}=0$, since a fraction with numerator $1$ can never equal zero). Hence Range $=\mathbb{R}-\{0\}$.

Check (independent verification). Substitute a test value: $x=4$ gives $f(4)=\dfrac{1}{4-3}=1$, consistent with $4$ being in the domain and $1$ being in the range. Try to solve $\dfrac{1}{x-3}=0$ directly: this would require $1=0$, which is impossible — confirming $0$ is correctly excluded from the range.

A common slip is to exclude the wrong value from the domain (e.g. excluding $x=0$ instead of $x=3$) by misreading the denominator. Another is to assume the range is 'all real numbers' without checking whether some value is genuinely unreachable — for a simple reciprocal-type function, the value that makes the numerator (here, $1$) impossible to achieve is exactly the value the vertical shift of $y=1/x$ leaves out.