434 CHAPTER 4 Inverse, Exponential, and Logarithmic Functions One-to-One Functions Suppose we define the following function F. F = 51-2, 22, 1-1, 12, 10, 02, 11, 32, 12, 526 (We have defined F so that each second component is used only once.) We can form another set of ordered pairs from F by interchanging the x- and y-values of each pair in F. We call this set G. G= 512, -22, 11, -12, 10, 02, 13, 12, 15, 226 G is the inverse of F. Function F was defined with each second component used only once, so set G will also be a function. (Each first component must be used only once.) In order for a function to have an inverse that is also a function, it must exhibit this one-to-one relationship. In a one-to-one function, each x-value corresponds to only one y-value, and each y-value corresponds to only one x-value. The function ƒ shown in Figure 1(a) is not one-to-one because the y-value 7 corresponds to two x-values, 2 and 3. That is, the ordered pairs 12, 72 and 13, 72 both belong to the function. The function ƒ in Figure 1(b) is one-to-one. 4.1 Inverse Functions ■ One-to-One Functions ■ Inverse Functions ■ Equations of Inverses ■ An Application of Inverse Functions to Cryptography Domain 1 f 2 3 4 5 6 7 8 9 Range Not One-to-One 1 6 Domain Range 7 8 5 2 3 4 One-to-One f (b) Figure 1 (a) One-to-One Function A function ƒ is a one-to-one function if, for elements a and b in the domain of ƒ, a 3b implies ƒ1a2 3ƒ1b2. That is, different values of the domain correspond to different values of the range. Using the concept of the contrapositive from the study of logic, the boldface statement in the preceding box is equivalent to ƒ1a2 =ƒ1b2 implies a =b. This means that if two range values are equal, then their corresponding domain values are equal. We use this statement to show that a function ƒ is one-to-one in Example 1(a). EXAMPLE 1 Deciding Whether Functions Are One-to-One Determine whether each function is one-to-one. (a) ƒ1x2 = -4x + 12 (b) ƒ1x2 = 225 - x2 SOLUTION (a) We can determine that the function ƒ1x2 = -4x + 12 is one-to-one by showing that ƒ1a2 = ƒ1b2 leads to the result a = b. ƒ1a2 = ƒ1b2 -4a + 12 = -4b + 12 ƒ1x2 = -4x + 12 -4a = -4b Subtract 12. a = b Divide by -4. By the definition, ƒ1x2 = -4x + 12 is one-to-one.
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