48
Chapter 1 | Functions and Graphs
Example 1.16 Finding Domain and Range for Algebraic Functions For each of the following functions, find the domain and range. a. f ( x ) = 3 x −1 5 x +2
b. To find the domain of f , we need 4 – x 2 ≥0 . Or, 4≥ x 2 Or x 2 ≤4 , the solution to which is – 2≤ x ≤2 . Therefore, the domain is { x | – 2≤ x ≤2} . If – 2≤ x ≤2 , then 0≤4 – x 2 ≤4 . Therefore, 0≤ 4 – x 2 ≤2 and the range of f is ⎧ ⎩ ⎨ y | 0≤ x ≤2 ⎫ ⎭ ⎬ . Solution a. It is not possible to divide by zero, so the domain is the set of real numbers x such that x ≠−2/5. To find the range, we need to find the values y for which there exists a real number x such that y = 3 x −1 5 x +2 . When we multiply both sides of this equation by 5 x +2, we see that x must satisfy the equation 5 xy +2 y =3 x −1.
From this equation, we can see that x must satisfy
2 y +1= x (3−5 y ).
If y =3/5, this equation has no solution. On the other hand, as long as y ≠3/5, x = 2 y +1 3−5 y
⎨ y | y ≠3/5 ⎫ ⎭ ⎬ .
satisfies this equation. We can conclude that the range of f is ⎧ ⎩
f , we need 4− x 2 ≥0. When we factor, we write 4− x 2 = (2− x )(2+ x ) ≥0. This inequality holds if and only if both terms are positive or both terms are negative. For both terms to be positive, we need to find x such that 2− x ≥0 and 2+ x ≥0. These two inequalities reduce to 2≥ x and x ≥−2. Therefore, the set { x | −2≤ x ≤2} must be part of the domain. For both terms to be negative, we need 2− x ≤0 and 2+ x ≥0. These two inequalities also reduce to 2≤ x and x ≥−2. There are no values of x that satisfy both of these inequalities. Thus, we can conclude the domain of this function is { x | −2≤ x ≤2}. If −2≤ x ≤2, then 0≤4− x 2 ≤4. Therefore, 0≤ 4− x 2 ≤2, and the range of f is
b. To find the domain of
⎧ ⎩ ⎨ y | 0≤ y ≤2
⎫ ⎭ ⎬ .
This OpenStax book is available for free at http://cnx.org/content/col11964/1.12
Made with FlippingBook - professional solution for displaying marketing and sales documents online