Introduction to Domain and Range
The domain of a function from set A to set B is all of set A. The range is either all or part of set B. In our example at the beginning of the chapter, we had A = {1, 2, 3,4}, B = {a, b, c} and our function was {(1, a ), (2, a ), (3, b), (4, b)}. The domain of this function is {1, 2, 3, 4}, and the range is all of the elements from B that were paired with elements from A. These were {a, b}.
For the functions in this book, the domain will consist of all the real numbers we are allowed to use for x . The range will be all of the y values. In this chapter, we will find the domain algebraically. In Chapter 3, we will find both the domain and range from graphs of functions.
Very often, we find the domain of a function by thinking about what we cannot do. For now the things we cannot do are limited to division by zero and taking even roots of negative numbers. If a function has x in a denominator, set the denominator equal to zero and solve for x. The domain will not include the solution(s) to this equation (assuming the equation has a solution). If a function has x under an even root sign, set the quantity under the sign greater than or equal to zero to find the domain. Later when we learn about logarithm functions and functions from trigonometry, we will have other things we cannot do. The domain and range are usually given in interval notation. There is a review of interval notation in the Appendix.
Examples

We cannot let x + 3 to be zero, so we cannot let x = −3. The domain is x ≠ −3, or (−∞, −3) ∪ (−3, ∞).

We will use factoring by grouping to factor the denominator.
x^{3} + 2 x^{2} − 2 = 0
x^{2}(x + 2) − 1(x + 2) = 0
( x + 2)(x^{2} − 1) = 0
(x + 2)(x − 1)(x + 1) = 0
x + 2 = 0 x − 1 = 0 x + 1 = 0
x = −2 x = 1 x = −1
The domain is all real numbers except 1, −1, and −2. The domain is shaded on the number line in Figure 2.1.

Because x^{2} + 1 = 0 has no real number solution, we can let x equal any real number. The domain is all real numbers, or (−∞, ∞).

Because we can only take the square root of nonnegative numbers, x − 8 must be nonnegative. We represent “ x − 8 must be nonnegative” as “ x − 8 ≥ 0.” Solving x − 8 ≥ 0, we get x ≥ 8. The domain is x ≥ 8, or [8, ∞).

We need to solve x^{2} − x − 2 ≥ 0. Factoring x^{2} − x − 2, we have ( x − 2)(x + 1).
x − 2 = 0 x + 1 = 0
x = 2 x = −1
Fig. 2.2
We will use x = −2 for the number to the left of −1, x = 0 for the number between −1 and 2, and x = 3 for the number to the right of 2 in x ^{2} − x −2 ≥ 0 to see which of these numbers makes it true.
Is (−2)^{2} − (−2) − 2 ≥ 0? Yes. Put “True” to the left of −1. Is 0^{2} − 0 − 2 ≥ 0? No. Put “False” between −1 and 2. Is 3^{2} − 3 − 2 ≥ 0? Yes. Put “True” to the right of 2.
Fig. 2.3.
The inequality is true for x ≤ −1 and x ≥ 2, so the domain is (−∞, −1] ∪ [2, ∞ ).

Because x ^{2} + 5 is always positive, we can let x be any real number. The domain is (−∞, ∞).

We can take the cube root of any number, so the domain is all real numbers, or (−∞, ∞).

f ( x ) = x^{4} − x^{2} + 1
There is no x in a denominator and no x under an even root sign, so the domain is all real numbers, or (−∞, ∞).
There are some functions that have x in a denominator and under an even root. At times, it will be useful to shade a number line to keep track of the domain.

We cannot let be zero, and we cannot let 4 − x be negative. These restrictions mean that we must have 4 − x > 0 (instead of 4 − x ≥ 0). The domain is 4 > x (or x < 4), which is the interval (−∞, 4).

Fig. 2.4
We also need for x ^{2} + 3 x − 4 not to be zero.
x ^{2} + 3 x − 4 = 0
(x + 4)(x − 1) = 0
x + 4 = 0 x − 1 = 0
x = −4 x = 1
We cannot let x = −4 and x = 1, so we will remove these numbers from x ≥ −10. The domain is [−10, −4) ∪ (−4, 1) ∪ (1, ∞).
Fig. 2.1.
The domain is (−∞, −2) ∪ (−2, −1) ∪ (−1, 1) ∪ (1, ∞).
At times the domain of a function will matter when we are solving an applied problem. For example, suppose there is a 10″ × 18″ piece of cardboard that will be made into an opentopped box. After cutting a square x by x inches from each corner, the sides will be folded up to form the box.
The volume of the box is a function of x, V ( x ) = x (18 − 2 x )(10 − 2 x ). What is the domain of this function? We obviously cannot cut a negative number of inches from each corner. If we cut 0 inches from each corner, we do not have a box, so x must be positive. Finally, the box is only 10 inches wide, so we can cut up to five inches from each corner. These facts make the domain 0 < x < 5. Maximizing the volume of this box is a typical problem in a first semester of calculus. The solutions to the mathematical problem are (approximately 2.0635, and 7.27008). Only one of these numbers is in the domain of the applied function, so only one of these numbers is the solution.
Find practice problems and solutions for these concepts at Domain and Range Practice Problems.
More practice problems for this concept can be found at: Introduction to Functions Practice Test.
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