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Home , Domain of a function, Range of a function

Lecture 2: Domain and range of functions

Victoria LEBED, [email protected]

MA1S11A: Calculus with Applications for Scientists

October 3, 2017 Yesterday we saw examples of functions X defined on certain parts of the real line R only—e.g., the root x √x, defined for x > 0; 7→ X defined by dierent formulas on dierent parts of R—e.g., the absolute value x |x|: 7→ x,x > 0, |x| = −x, x < 0.

These parts are typically intervals, rays, and various combinations thereof. Today we will X learn how to handle these intervals, rays, etc. mathematically (this will be our tools), X discuss how to determine for which values of the independent variable a is defined, and what values it takes (this is where we will apply our tools). 1 Reminder: intervals

[a, b] stands for the closed interval consisting of all x for which a 6 x 6 b. Mathematically, this can be wrien in a very eicient way: [a, b]= { x R | a 6 x 6 b }. ∈ Read: [a, b] is the of all real values x such that a 6 x 6 b holds. A more “human” reading: [a, b] is the set of all real x between a and b.

Similarly, the open interval (a, b) is defined by (a, b)= { x R | a

Half-closed intervals: [a, b) = { x R | a 6 x

An open end of an interval may assume infinite values, in which case an interval becomes a ray: for example, [a, + )= { x R | a 6 x }, ∈ is closed ray, while (− ∞,b)= { x R | x

Alice and Bob are flatmates. Yesterday Alice was at home from 6am to midday, and Bob from 1am to 7am.

estion 1. When were both of them at home? Solution. Consider the intervals A = [6, 12], and B = [1, 7]. The question asks to compute the intersection of these time intervals: A B = [6, 12] [1, 7] = [6, 7]. ∩ ∩ So, Alice and Bob were both at home from 6am to 7am.

estion 2. When was at least one of them at home? Solution. We need to compute the union of our time intervals: A B = [6, 12] [1, 7] = [1, 12]. ∪ ∪ 3 Operations on intervals: example

Alice and Bob are flatmates. Yesterday Alice was at home from 6am to midday, and Bob from 1am to 7am.

estion 3. When was none one of them at home? Solution. We will use one more interval D = [0, 24], encoding the whole day (yesterday). We need to compute the dierence of time intervals: D \ (A B) = [0, 24] \ [1, 12] = [0, 1] [12, 24]. ∪ ∪ Unlike in the two first questions, we obtained a part of R which is disconnected, and thus is neither an interval nor a ray. These are the more complicated parts of R referred to above. 4 Operations on intervals: definitions

Let A and B be two parts of R (e.g., intervals or rays). The intersection of A and B consists of those points that belong to both A and B: A B = { x R | x A and x B }. ∩ ∈ ∈ ∈

The union of A and B consists of those points that belong either to A or else to B: A B = { x R | x A or x B }. ∪ ∈ ∈ ∈ B This includes points belonging to both A and B. In maths, “or” is inclusive! The dierence of A and B consists of those points that belong to A but not to B: A\B = { x R | x A and x / B }. ∈ ∈ ∈

Remark. Similar definitions work for any sets. 5

The function x √x is an example of a function which is only 7→ defined for some values of the independent variable x, in this case for x > 0. Another example is the function x 1/x, in which case all values of x 7→ except for x = 0 are allowed. Definition. The domain of a function f consists of all values of x for which the value f(x) is defined. For a function defined by a table, its domain consists of numbers in the first row: x 1 1.5 2 2.5 3 3.5 4 y 0 1.25 3 5.25 8 11.25 15 5 Domain of a function

For a function defined by a graph, its domain consists of the values for which the corresponding vertical line meets the graph: y 3 2 1 x −3 −2 −1 1 2 3 −1 −2 −3 In other words, the domain is the projection of the graph on the x-axis. estion. What is the domain of the function defined by the graph above? Answer. [−1.5, 1.5]. 6 Natural domain of a function

With the expressions like √x or 1/x, we saw how the domain of a function may be restricted by a formula from which we compute its values.

Definition. The natural domain of a function f defined by a formula consists of all values of x for which f(x) has a well defined real value.

Example 1. The natural domain of f(x)= x3 consists of all real numbers (− , ), since for each real x its cube is a well defined .

∞ ∞ 7 Natural domain: examples

1 Example 2. The natural domain of f(x)= (x−1)(x−3) consists of all real numbers except for x = 1 and x = 3, since for those numbers division by zero occurs. One can write the answer as either { x R | x = 1 and x = 3 }, ∈ 6 6 or, using the union , (− , 1) (1, 3) (3, + ). ∪ ∪

∞ ∞sin x Example 3. The natural domain of f(x)= tan x = cos x consists of all numbers except for those where cos x = 0. The cosine vanishes precisely at odd multiples of π/2, so the natural domain of tan is { x R | x = π/2, 3π/2, 5π/2, . . . } ∈ 6 ± ± ± = . . . (−3π/2, −π/2) (−π/2, π/2) (π/2, 3π/2) (3π/2, 5π/2) .... ∪ ∪ ∪ ∪ ∪ 7 Natural domain: examples

Example 4. The natural domain of f(x)= √x2 − 5x + 6 consists of all numbers x for which x2 − 5x + 6 > 0 (in order for the square root to assume a real value). Since x2 − 5x + 6 = (x − 2)(x − 3) (we use the 5 √52−4 6 1 formula for roots of a , x = ± 2 · · ), we would like to find x for which (x − 2)(x − 3) > 0.

7 y 6 5 4 3 2 1 x −−11 1 2 3 4

Thus, the natural domain of f is (− , 2] [3, + ). ∪ ∞ ∞ 7 Natural domain: examples

2 Example 5. The natural domain of f(x)= x −5x+6 consists of all x = 2 x−2 6 since the only reason for the value of f to be in this case is where division by zero occurs. However, since x2 − 5x + 6 = (x − 2)(x − 3), algebraically this expression can be simplified to x − 3. The laer expression is defined for x = 2 as well. B When performing algebraic operations, it is important to keep track of original domains! In this particular case we must write f(x)= x − 3 for x = 2 6 (and f(x) is undefined for x = 2). Thus we do not lose information on the original domain. y 2 1 x −−11 12345bc −2 −3 −4 8 Natural domain:

Aer all the examples above, the following algorithm for describing the natural domain of a basic function defined by a formula should appear intuitive to you: 1) Whenever you see an expression of type pg(x) (or 2npg(x) with n N), ∈ determine all x for which g(x) > 0. h(x) 2) Whenever you see an expression of type g(x) , determine all x for which g(x) = 0. 6 sin(x) cos(x) B Beware of the functions tan(x)= cos(x) and cot(x)= sin(x) , which are implicit fractions! 3) Determine the intersection of all the parts of R obtained in the previous points. 9

Let us look at the two functions we already examined when talking about domains: f(x)= 1/x and f(x)= √x. What values do those functions assume for the values of x ranging over their natural domains? The first function f(x)= 1/x assumes all nonzero values. Indeed, to aain the value v = 0, we should just observe that 1/(1/v)= v. 6 The second function f(x)= √x assumes all nonnegative values. Indeed, to aain the value v > 0, we should just observe that for such values we have √v2 = v. Definition. The range of a function f consists of all values f(x) it assumes when x ranges over its domain. For a function defined by a table, its range consists of numbers in the second row:

x 1 1.5 2 2.5 3 3.5 4 y 0 1.25 3 5.25 8 11.25 15 9 Range of a function

Definition. The range of a function f consists of all values f(x) it assumes when x ranges over its domain. For a function defined by a graph, its range is the projection of its graph on the y-axis: y 3 2 1 x −3 −2 −1 1 2 3 −1 −2 −3 The range of this function is [0, 1.5]. 9 Range of a function

Definition. The range of a function f consists of all values f(x) it assumes when x ranges over its domain.

Example 1. The range of f(x)= 2 + √x − 1 is [2, + ). To see that, we observe that the natural domain of this function is [1, + ) since we request that the expression from which we∞ extract the square root is non-negative. ∞ Now, as x increases from 1 to + , X x − 1 varies in [0, + ), ∞ X √x − 1 varies in [0, + ), X and 2 + √x − 1 varies∞ in [2, + ). ∞ ∞ 9 Range of a function

x+1 Example 2. The range of f(x)= x−1 is harder to guess, so we shall approach it directly from the definition. We would like to find the values of a for x+1 which the equation x−1 = a has solutions. Solving it, we get x + 1 = ax − a, a + 1 = ax − x = x(a − 1), a + 1 x = . a − 1 From x = a+1 , we see that all a R except for a = 1 are in the natural a−1 ∈ range. So, the natural range is (− , 1) (1, + ). ∪ x+1 B We should be careful when going from x−1 = a to x + 1 = a(x − 1) because we change the natural domain.∞ However,∞ x = 1 is not a solution to x + 1 = a(x − 1) anyway.