A radical function's domain is not read off — it is solved for. An even index turns the requirement "radicand ≥ 0" into an inequality, and the answer to that inequality is the domain. The graph then follows from the domain, the range and a handful of exact points.
Core ideas
The course ends where it began: with the values that are not allowed, found before anything is drawn.
For an even index, f(x) = ⁿ√(g(x)) is defined exactly where g(x) ≥ 0. Finding the domain is therefore an inequality problem from Linear Equations and Inequalities, and when g is a quadratic it is a sign-analysis problem from Polynomials and Factoring.
∛x is defined for every real number, negatives included, and its range is every real number too. The whole difficulty of this lesson is a feature of even indices only.
√(anything) is never negative, so the basic square-root function has range y ≥ 0 — it is half of a sideways parabola, not all of one. A vertical shift moves that floor; a reflection turns it into a ceiling.
The material
How to find the domain, what the graph looks like, and what the transformations do to both.
A radical function is one whose rule places the variable under a radical: f(x) = ⁿ√(g(x)). For even n its domain is { x : g(x) ≥ 0 }; for odd n it is the domain of g.
Compare this with “Rational Expressions and Their Domains”. There the domain was found by setting a denominator equal to zero and excluding the roots — an equation. Here it is found by requiring a radicand to be non-negative — an inequality. A denominator excludes isolated points; an even radical excludes whole intervals.
√(x − 3) needs x − 3 ≥ 0, so x ≥ 3. √(5 − x) needs 5 − x ≥ 0, so x ≤ 5 — the inequality reverses, exactly as in Linear Equations and Inequalities. ∛(x − 3) needs nothing: its domain is all real numbers.
For h(x) = √(x² − 9) the requirement is x² − 9 ≥ 0, that is (x − 3)(x + 3) ≥ 0. A product of two factors is non-negative when both have the same sign, which happens for x ≤ −3 and for x ≥ 3. The domain is two separate rays, and the graph has a gap between −3 and 3 where the function does not exist.
This is worth pausing on. The formula √(x² − 9) gives no warning that it is undefined on an interval six units wide. Only solving the inequality reveals it, and a plot made by sampling convenient integers would show the gap without explaining it.
y = √x has domain x ≥ 0 and range y ≥ 0. It starts at the origin, rises, and flattens: it is the upper half of the parabola x = y².
It is the reflection of y = x² restricted to x ≥ 0 in the line y = x, which is what makes it that function's inverse — and why the restriction is needed at all.
The transformations from Lines, Functions and Graphs apply unchanged. In y = a·√(x − h) + k the graph starts at (h, k) instead of the origin; h shifts the domain to x ≥ h, k shifts the range to y ≥ k, and a negative a flips it downwards so the range becomes y ≤ k. The endpoint is the one point worth plotting exactly.
One last connection back to “Simplifying Radical Expressions”. The function f(x) = √(x²) is defined for every real x, since x² is never negative — but its graph is not the line y = x. It is y = |x|, a V with its vertex at the origin. The identity from “Simplifying Radical Expressions”, drawn.
Everywhere in this course √(−4) has been outside the domain, and every domain here has been a set of real numbers. Quadratics and Complex Numbers introduces i and gives negative radicands values; nothing in this course anticipates that, and the domains stated here are the correct ones for real-valued functions.
Interactive
Where a square root refuses, computed from the radicand
Method
Domain first. Everything else is easier once it is known, and wrong if it is skipped.
Odd index: the domain is everything g allows, and you are done with this question. Even index: continue to the inequality.
A linear radicand gives a single ray — remember to reverse the inequality when dividing by a negative. A quadratic radicand needs factoring and a sign analysis, and may give two rays or an interval.
At a domain boundary caused by a zero radicand, evaluate the whole rule: a linear radicand usually gives one endpoint, while a quadratic may give two. For a√(x − h) + k, the principal root keeps the range on or above k when a > 0 and on or below k when a < 0.
Choose values that make the radicand a perfect power: for √(x − 3) take x = 3, 4, 7, 12 to get y = 0, 1, 2, 3. Four exact points beat twenty decimals.
One linear radicand, one quadratic, and then the curve for the first.
In (b) the two checks are not decoration. x = 0 confirms the gap is genuine rather than an artefact of the factoring, and x = 3 confirms the endpoints belong to the domain — ≥ rather than >, because √0 is perfectly well defined.
In (c) the x values were chosen so the radicand is 0, 1, 4, 9, 16. Picking x = 5 instead gives y = √2 ≈ 1.414, a point you cannot place accurately by hand. Choosing inputs that make the radicand a perfect power is the difference between a sketch and a guess.
For a faded graph analysis, use g(x) = −√(6 − 2x) + 1. The supplied condition is 6 − 2x ≥ 0. Solve it, then use the outside minus and vertical shift to state the endpoint and range before choosing exact points. Check against domain x ≤ 3, endpoint (3, 1), range y ≤ 1, and points (1, −1) and (−5, −3). A domain x ≥ 3 means the inequality was divided by −2 without reversing it; a range y ≥ 1 means the reflection was applied to the graph but not its outputs.
Given any radical function you should be able to decide the domain by solving an inequality, state the range from the principal-root convention, locate the endpoint exactly, and sketch the curve from points you chose to be exact. A graph drawn before the domain is known is a graph that will be drawn over an interval where the function does not exist.
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