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Irrational (Radical) Functions

Radical expressions often encode geometric lengths and appear in limits, derivatives, and integrals.

Domain reminder: the radicand of an even root must be nonnegative; an odd root accepts any real number.

Definition

$f(x) = \sqrt[n]{g(x)}$ where $n$ is a positive integer and $g(x)$ is another function.

Symbol	Type	Pronunciation/Note	Meaning in this article
$\sqrt[n]{g(x)}$	Mathematical symbol	n-th root of g of x	General form of radical function
$\mathbb{R}$	Mathematical symbol	Real numbers	Set of real numbers being discussed

Even $n$ : require $g(x) \ge 0$ for real outputs.
Odd $n$ : domain may be all real numbers.

Typical Shapes

$f(x) = \sqrt{x}$ : domain $[0,+\infty)$ , increasing, concave down.
$f(x) = \sqrt{4 - x^2}$ : domain $[-2,2]$ , upper semicircle.
$f(x) = \sqrt[3]{x}$ : domain $\mathbb{R}$ , odd symmetry.

Basic Properties

Domain

For $f(x) = \sqrt[n]{g(x)}$ the domain is decided entirely by the radicand $g(x)$ :

If $n$ is even, the $n$ -th root exists only for a nonnegative radicand, so the domain is $\{x \mid g(x) \geq 0\}$ .
If $n$ is odd, every real radicand has a real $n$ -th root, so the domain is $\{x \mid g(x) \in \mathbb{R}\}$ , that is, the domain of $g$ itself.

Monotonicity

If $g(x)$ is increasing and $g(x) \geq 0$ on an interval, then $f(x) = \sqrt[n]{g(x)}$ is increasing on that interval.
If $g(x)$ is decreasing and $g(x) \geq 0$ on an interval, then $f(x)$ is decreasing on that interval.

The reason is that the map $u \mapsto \sqrt[n]{u}$ is increasing wherever it is defined — on $[0,+\infty)$ for even $n$ and on all of $\mathbb{R}$ for odd $n$ — so it preserves the order of the values of $g$ .

Parity

For $f(x) = \sqrt[n]{x}$ :

If $n$ is even, the domain $[0,+\infty)$ is not symmetric about the origin, so parity cannot even be discussed and the function is neither even nor odd.
If $n$ is odd, the domain $\mathbb{R}$ is symmetric about the origin and $f(-x) = \sqrt[n]{-x} = -\sqrt[n]{x} = -f(x)$ , so the function is odd.

Techniques

Rationalization: multiply by conjugates to simplify limits or equations.
Domain checks: combine conditions when radicals appear in fractions.
Composition: $\sqrt{g(x)}$ inherits domain constraints from $g$ .

Worked Examples

Example 1: Domain of a square-root function

Determine the domain of the radical function $f(x) = \sqrt{x^2 - 4}$ .

Reference Answer

irrational functiondomainsquare root function

Approach: The radicand of an even root must be nonnegative, so the domain is exactly the solution set of the inequality $x^2 - 4 \geq 0$ .

Steps:

Identify the radicand: it is $x^2 - 4$ .
Write down the requirement $x^2 - 4 \geq 0$ .
Solve it: $x^2 \geq 4$ , which gives $x \leq -2$ or $x \geq 2$ .
Both endpoints are included, because a radicand equal to zero still has a real square root.

Answer:

The domain is $(-\infty, -2] \cup [2, +\infty)$ .

Example 2: Domain of a cube-root function

Determine the domain of the cube-root function $f(x) = \sqrt[3]{x^3 - 8}$ .

Reference Answer

irrational functiondomaincube root function

Approach: An odd root accepts every real radicand, so the domain of $f$ is simply the domain of the inner expression.

Steps:

Identify the radicand: it is $x^3 - 8$ .
A cube root is defined for every real number, so the only requirement is that $x^3 - 8$ itself be a real number.
The polynomial $x^3 - 8$ is defined for every real $x$ ; in particular the radicand may vanish, for instance $f(2) = \sqrt[3]{0} = 0$ , which causes no problem.

Answer:

The domain is $(-\infty, +\infty)$ ; there is no restriction on $x$ .

Exercises

Exercise 1

Find the domain of $f(x) = \dfrac{\sqrt{x-1}}{x-3}$ .

Reference Answer

irrational functiondomainradical function

Approach:

For a quotient with a radical in the numerator, the radicand must be non-negative and the denominator must not vanish; combine the two restrictions.

Steps:

The radical $\sqrt{x-1}$ requires $x - 1 \geq 0$ , that is $x \geq 1$ .
The denominator requires $x - 3 \neq 0$ , that is $x \neq 3$ .
Combining them leaves $[1,3) \cup (3,+\infty)$ .

Answer:

Domain: $[1,3) \cup (3,+\infty)$ .

Exercise 2

Solve $\sqrt{x+5} = x - 1$ in $\mathbb{R}$ .

Reference Answer

irrational functionradical equationdomain restrictions

Approach:

Squaring a radical equation is only valid under the sign condition it creates, so record the domain restriction first, then square and finally check every candidate against it.

Steps:

The equation $\sqrt{x+5} = x - 1$ requires the right-hand side to be non-negative, so $x - 1 \geq 0$ , that is $x \geq 1$ .
Squaring both sides gives $x + 5 = (x-1)^2 = x^2 - 2x + 1$ , hence $x^2 - 3x - 4 = 0$ and $(x-4)(x+1) = 0$ .
The candidates are $x = 4$ or $x = -1$ .
Checking the domain, only $x = 4$ satisfies $x \geq 1$ .

Answer:

Require $x-1 \ge 0$ ⇒ $x \ge 1$ . Square both sides: $x + 5 = (x-1)^2 = x^2 - 2x + 1$ ⇒ $x^2 - 3x - 4 = 0$ ⇒ $(x-4)(x+1)=0$ ⇒ candidates $x=4$ or $x=-1$ . Check domain: only $x=4$ is valid.

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation/Explanation	Meaning in This Article
$\sqrt[n]{g(x)}$	Mathematical symbol	n-th root of g of x	The general form of a radical function
$[a,b]$	Mathematical symbol	closed interval a b	The interval notation of a sample domain
$\mathbb{R}$	Mathematical symbol	Real numbers	The number set under discussion

Rational Functions

Function Exercises

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Irrational (Radical) Functions

Definition

Typical Shapes

Basic Properties

Domain

Monotonicity

Parity

Techniques

Worked Examples

Example 1: Domain of a square-root function

Example 2: Domain of a cube-root function

Exercises

Exercise 1

Exercise 2

Summary

Symbols Used in This Article

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