Domain and Range

Domain. Look for inputs that would make the formula undefined:

• Division by zero. $f(x) = \dfrac{1}{x - 2}$ has domain $x \ne 2$.
• Square root of negative. $f(x) = \sqrt{x - 3}$ requires $x - 3 \ge 0$, so $x \ge 3$.
• Logarithm of non-positive. $f(x) = \log(x)$ requires $x > 0$.

If no restriction applies, the domain is ALL REALS ($\mathbb{R}$).

Worked examples:

Function	Domain
$f(x) = 2x + 7$	$\mathbb{R}$ (no restrictions)
$f(x) = \sqrt{x - 4}$	$x \ge 4$
$f(x) = \dfrac{1}{x + 5}$	$x \ne -5$
$f(x) = \dfrac{\sqrt{x}}{x - 3}$	$x \ge 0$ AND $x \ne 3$
$f(x) = \log(x - 1)$	$x > 1$

Range. Reason about what outputs are reachable:

• $f(x) = x^2$: $x^2 \ge 0$ always, so range is $y \ge 0$.
• $f(x) = -x^2 + 4$: max value 4 (at $x = 0$), so range is $y \le 4$.
• $f(x) = \sqrt{x}$: outputs are $y \ge 0$.
• $f(x) = \dfrac{1}{x}$: outputs are $y \ne 0$ (the function never equals zero).

For a more complex function, sketch the graph and read off the range.

Domain = horizontal extent of the graph (project shadow onto x-axis).

Range = vertical extent (project shadow onto y-axis).

Worked example. A parabola $y = x^2 - 4$.

• Domain: $\mathbb{R}$ (the parabola exists for all $x$).
• Range: minimum at vertex $(0, -4)$, opens upward → $y \ge -4$.

Worked example. A graph that starts at $(2, 0)$ and rises to $(5, 6)$ (e.g. a line segment).

• Domain: $2 \le x \le 5$ (or $[2, 5]$).
• Range: $0 \le y \le 6$ (or $[0, 6]$).

Interval notation for domain/range:

• $[a, b]$: closed interval, both endpoints included.
• $(a, b)$: open interval.
• $(-\infty, b]$: 'all $x$ up to and including $b$'.
• $[a, \infty)$: 'all $x$ from $a$ upwards'.
• $\mathbb{R}$ or $(-\infty, \infty)$: all reals.

Restricted domains. Sometimes a function is DEFINED on a restricted domain — e.g. 'the height of a basketball is $h(t) = -5t^2 + 20t + 1$ for $0 \le t \le t_{\text{ground}}$'. In real contexts, ALWAYS check whether the domain has a physical restriction.