In and out.
domain
x
f →
range
y
domain
every x you are allowed to feed in
range
every y that can come out
Start by assuming every real number is allowed, then look for the two things that ban some of them.
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Domain and Range
Which inputs are allowed, and which outputs can actually come out.
Find the domain
Domain of f(x) = √x − 3
A square root needs its inside to be zero or positive: x − 3 ≥ 0, so x ≥ 3. B wrongly excludes 3 — √0 = 0 is perfectly legal. Only a denominator bans equality; a root allows it.
The only two rules you need.
never divide by zero
set the bottom ≠ 0
never square-root a negative
set the inside ≥ 0
Any function with no fraction and no even root has every real number as its domain — polynomials, for instance.
A cube root is fine with negatives: 3√−8 = −2. Only even roots are fussy about the sign.
Apply them to the two functions.
√x − 3 needs x − 3 ≥ 0
the root rule
x ≥ 3
or [3, ∞)
1 ÷ (x − 2) needs x − 2 ≠ 0
the fraction rule
x ≠ 2
everything else is fine
≥ for roots, ≠ for fractions
A root is happy at zero — √0 = 0. A denominator is not. Writing x > 3 instead of x ≥ 3 loses a legitimate value.
Range: think about the shape.
y = x² + 1 is a parabola that opens upward.
x² ≥ 0 for every x
a square is never negative
x² + 1 ≥ 1
add one to both sides
y ≥ 1
or [1, ∞)
The vertex is at (0, 1) and the curve only goes up from there. For a parabola, the range always starts at the vertex.
Any parabola: find the vertex, then look which way it opens.
Find the range of y = −(x − 1)² + 4, and of y = 3x² − 12x + 5.
−(x − 1)² ≤ 0
a square made negative is never positive
y ≤ 4
opens down: the vertex (1, 4) is the top
3x² − 12x + 5: x = −b ÷ 2a = 12 ÷ 6 = 2
not in vertex form — find the vertex's x first
y = 3(4) − 12(2) + 5 = 12 − 24 + 5 = −7
put it back in for the vertex's y
y ≥ −7
a = 3 > 0 opens up: the vertex is the bottom
The constant term is not the vertex
In 3x² − 12x + 5 the 5 is only the y-intercept, so “y ≥ 5” is wrong. And in −(x − 1)² + 4 the 1 is an x-value; the range is about y, so it is y ≤ 4.
Terms out of order? Put them in standard order first. In y = 1 + 6x − x², a = −1 and b = 6: x = −6 ÷ (−2) = 3, y = 1 + 18 − 9 = 10, and a < 0 opens down, so y ≤ 10.
Both rules at once: a root over a fraction.
Find the domain of f(x) = √5 − x ÷ (x + 1).
5 − x ≥ 0, so x ≤ 5
the root rule — 5 is allowed
x + 1 ≠ 0, so x ≠ −1
the fraction rule
(−∞, −1) ∪ (−1, 5]
x ≤ 5 with one hole punched at −1
Draw the number line: shade everything up to 5 (closed dot), then open a hole at −1. The ∪ joins the two pieces either side of the hole. If the hole had fallen outside the shaded part, it would change nothing. A denominator like x² − x − 12 is factored first — (x − 4)(x + 3) — and every zero of it is a hole.
Two roots added, like √x + 1 + √4 − x: each inside must be ≥ 0, so x ≥ −1 AND x ≤ 4, which is −1 ≤ x ≤ 4. Asked how many integers? Count both ends: −1, 0, 1, 2, 3, 4 is 6 integers.
Variations: a root underneath, a minus in front of x, a range with a gap.
A root in the denominator. Find the domain of f(x) = 3 ÷ √x + 5.
x + 5 ≥ 0
the root rule
√x + 5 ≠ 0, so x + 5 ≠ 0
the fraction rule on the same expression
x + 5 > 0, so x > −5
or (−5, ∞)
What is different: both rules hit one expression, so ≥ tightens to > and the bracket at −5 is round.
A minus in front of x. Find the domain of f(x) = √10 − 4x.
10 − 4x ≥ 0
the root rule
−4x ≥ −10
take 10 from both sides
x ≤ 2.5
divide by −4: the sign flips
What is different: the allowed side is to the LEFT of the endpoint. Check x = 0: √10 is fine, and 0 ≤ 2.5 agrees.
Forgetting the flip
Writing x ≥ 2.5 keeps the wrong side. Test x = 3: √10 − 12 = √−2, not a real number, so the answer cannot be x ≥ 2.5.
A range that is not a parabola. Find the range of y = 1 ÷ (x − 4) + 2.
1 ÷ (x − 4) ≠ 0 for every x
a fraction with top 1 is never zero
1 ÷ (x − 4) + 2 ≠ 2
add 2 to both sides
y ≠ 2, while x ≠ 4
range and domain each lose one value
What is different: the banned output comes from the added constant, not the denominator. The 4 belongs to the domain; the 2 belongs to the range.
A root shifted up or down. Find the range of y = √x + 1 − 3.
√x + 1 ≥ 0
a square root is never negative
√x + 1 − 3 ≥ −3, so y ≥ −3
subtract 3 from both sides
What is different: the number inside the root moves the domain only; the number outside moves the range.
The common ones, at a glance.
| Function | Domain | Range |
|---|---|---|
| y = 2x + 5 | all real | all real |
| y = x² | all real | y ≥ 0 |
| y = √x | x ≥ 0 | y ≥ 0 |
| y = 1x | x ≠ 0 | y ≠ 0 |
| y = |x| | all real | y ≥ 0 |
| y = x³ | all real | all real |
[ ] square bracket
the endpoint is included
( ) round bracket
excluded — and ∞ always gets one
On the exam
Expect a fraction whose bottom must be factored, a root over a fraction (answer in ∪ notation), or the range of a parabola. For a range, find the vertex and the sign of a — 10 seconds. Then check the brackets: [ ] where a root allows 0, ( ) at every hole and at ∞.
Check: test a legal and an illegal value.
√x − 3 at x = 3
√0 = 0
allowed ✓
at x = 2
√−1
not a real number ✗
1 ÷ (x − 2) at x = 2
1 ÷ 0
undefined ✗
x² + 1 at x = 0
1
the smallest output ✓
Now you try
Practice problem
Find the domain
f(x) = √2x + 8
Set the inside at or above zero: 2x + 8 ≥ 0 gives 2x ≥ −8, so x ≥ −4. D forgets to divide by the 2. Dividing by a positive never flips the inequality, so the sign stays as it is.