Domain and Range

What may go in, and what can come out

The domain of a function is the set of inputs it accepts. The range is the set of outputs it actually produces.

Take $f\left(x\right)=x^2$f(x) = x². Any number at all can be squared: positive, negative, zero, whole or not. So every real number is allowed in, and the domain of f is all real numbers.

The graph of $y=x^2$y = x². It carries on to the left and to the right without end, because every x has a square.

The range of $x^2$x²

A square is never negative. A positive number times itself is positive, a negative number times itself is also positive, and $0^2=0$0² = 0. So no output of $x^2$x² is below 0.

Every other output is reached. 0 comes from x = 0, and any positive number comes from its square root: 9 from x = 3 or x = −3, and 2 from $x=\sqrt{2}$x = √2. So the range of $x^2$x² is every number from 0 upward, written $f\left(x\right)\ge 0$f(x) ≥ 0. On the graph, the curve never goes below the x-axis, and it touches it at the origin.

Adding a number moves the whole range. For $f\left(x\right)=x^2+5$f(x) = x² + 5, the smallest $x^2$x² is 0, so the smallest output is 0 + 5 = 5, and the range is $f\left(x\right)\ge 5$f(x) ≥ 5.

Division by zero

For $f\left(x\right)=\frac{1}{x}$f(x) = 1/x, the input x = 0 would mean dividing by 0, which has no answer. Every other number is allowed, so the domain is all real numbers except 0, written $x\ne 0$x ≠ 0.

The output is never 0 either, because 1 divided by any number is not 0. Every other output is reached: 5 comes from $x=\frac{1}{5}$x = 1/5, and −2 from $x=-\frac{1}{2}$x = −1/2. So the range is $f\left(x\right)\ne 0$f(x) ≠ 0.

Find the forbidden input by asking what makes the denominator 0. For $f\left(x\right)=\frac{1}{\left(x-6\right)}$f(x) = 1/(x − 6), the denominator x − 6 is 0 when x = 6, so the domain is $x\ne 6$x ≠ 6. The input 0 is allowed there: it gives $\frac{1}{\left(-6\right)}$1/(−6).

The graph of $y=\frac{1}{x}$y = 1/x comes in two pieces, one on each side of the y-axis, which is the line x = 0. No point of the graph has x = 0, and none has y = 0.

Square roots

No real number squares to a negative number, so a negative number has no real square root. For $f\left(x\right)=\sqrt{x}$f(x) = √x, the input must be 0 or more: the domain is $x\ge 0$x ≥ 0. The square root sign means the root that is not negative, so the range is $f\left(x\right)\ge 0$f(x) ≥ 0 as well.

What has to be 0 or more is the whole expression inside the root. For $f\left(x\right)=\sqrt{x-5}$f(x) = √(x − 5), $x-5\ge 0$x − 5 ≥ 0, so $x\ge 5$x ≥ 5. The smallest input is 5, where the output is $\sqrt{0}=0$√0 = 0. The input 0 is not allowed: it puts −5 under the root.

The gold band on the x-axis is the domain and the green band on the y-axis is the range. A ring marks a value that is left out. The figure writes all real numbers as ℝ. For every number from 0 upward it writes an interval: a square bracket before the 0, which means 0 is included, then $\infty$∞ with a round bracket, because the numbers go on without end. Choose $\sqrt{x}$√x and move the point left: it stops at x = 0, where the domain begins.

Start from every number, then remove

To find a domain, start with all real numbers and remove every input that breaks a rule: a denominator of 0, or a negative number under a square root.

For $f\left(x\right)=\sqrt{x-3}+1$f(x) = √(x − 3) + 1, the root needs $x-3\ge 0$x − 3 ≥ 0, so the domain is $x\ge 3$x ≥ 3. The smallest the root can be is 0, at x = 3, so the smallest output is 0 + 1 = 1, and the range is $f\left(x\right)\ge 1$f(x) ≥ 1. Test a value: x = 7 gives $\sqrt{4}+1=3$√4 + 1 = 3, which is in the range.

For $f\left(x\right)=4-x^2$f(x) = 4 − x², every input is allowed. $x^2$x² is never negative, so $4-x^2$4 − x² is never more than 4, and it equals 4 at x = 0. The range is $f\left(x\right)\le 4$f(x) ≤ 4.

Three more functions. $4-x^2$4 − x² accepts every input, and its range stops at 4. $\frac{1}{\left(x-1\right)}$1/(x − 1) leaves out x = 1 and the output 0. Choose $\sqrt{x+3}$√(x + 3) and move the point left: it stops at x = −3, where x + 3 = 0.

When the domain is cut short

A real setting can make the domain smaller than the rule allows: a length must be positive, and a number of people cannot be negative. The range is then the set of outputs over that smaller domain.

Take $f\left(x\right)=x^2$f(x) = x² with the domain $-1\le x\le 3$−1 ≤ x ≤ 3. At the ends, f(−1) = 1 and f(3) = 9. But x = 0 is inside the domain too, and f(0) = 0 is smaller than either end. So the range is $0\le f\left(x\right)\le 9$0 ≤ f(x) ≤ 9, not $1\le f\left(x\right)\le 9$1 ≤ f(x) ≤ 9. Look at the ends of the domain and at any turning point inside it.

The usual mistakes

Forgetting the added number. The smallest output of $x^2+5$x² + 5 is not 0: the smallest $x^2$x² is 0, and then 5 is added, so it is 5.

Excluding the wrong input. For $\frac{1}{\left(x-6\right)}$1/(x − 6), x = 0 gives $\frac{1}{\left(-6\right)}$1/(−6), which is fine, and x = −6 gives $\frac{1}{\left(-12\right)}$1/(−12). Only x = 6 makes the denominator 0.

Starting a root’s domain at 0. For $\sqrt{x-5}$√(x − 5), x = 0 puts −5 under the root. The domain starts where the inside is 0, at x = 5.

Finding a range from the ends of the domain alone. A turning point inside the domain can give a smaller or larger output than either end.