Domain of definition: the forbidden-values checklist

The idea

The domain $D_f$ is the set of $x$ for which $f(x)$ exists.

The forbidden-values checklist

For every expression, exclude:

Expression	Forbidden
$\dfrac{A}{B}$	$B = 0$
$\sqrt{A}$	$A < 0$
$\ln(A)$	$A \leq 0$
$\tan(x)$	$x = \pi/2 + k\pi$
$A^\alpha$ non-integer	$A \leq 0$

Method

1. Identify each sensitive expression in $f(x)$.
2. Write the conditions.
3. Solve the system of conditions.
4. Write $D_f$ as an intersection (or union of intervals).

Examples

$f(x) = \dfrac{1}{x - 3}$ ⇒ $x \neq 3$. $D_f = \mathbb{R} \setminus \{3\}$.

$f(x) = \sqrt{x - 5}$ ⇒ $x - 5 \geq 0$ ⇒ $x \geq 5$. $D_f = [5, +\infty)$.

$f(x) = \ln(x + 2)$ ⇒ $x + 2 > 0$ ⇒ $x > -2$. $D_f = (-2, +\infty)$.

$f(x) = \dfrac{\sqrt{x}}{x - 4}$ ⇒ $x \geq 0$ AND $x \neq 4$. $D_f = [0, 4) \cup (4, +\infty)$.

Composition: careful

$f(x) = \ln(x^2 - 4)$ ⇒ $x^2 - 4 > 0$ ⇒ $x < -2$ or $x > 2$. $D_f = (-\infty, -2) \cup (2, +\infty)$.

Solve the inequality on the argument, don’t just state it!

Note

• A polynomial is defined on $\mathbb{R}$ (no restriction).
• $\sqrt{A}$ requires $A \geq 0$ (non-strict: $\sqrt{0} = 0$ exists).
• $\ln(A)$ requires $A > 0$ (strict: $\ln(0)$ doesn’t exist).

Common mistakes

1. Confusing $\geq$ (root) and $>$ (log).
2. Forgetting to solve the inequality (writing the condition isn’t enough).
3. Thinking $\dfrac{1}{x}$ is defined at 0 (forbidden).
4. Writing $D_f = \mathbb{R}$ without checking.

?À toi de jouer

Domain of f(x) = √(2x − 6)?

A.[3, +∞)B.(3, +∞)C.ℝD.(−∞, 3]

Valider