1. Properties of Functions
Section 4.3
Properties of Functions

2. Activity 1 Define One to one functions Onto functions
Invjective functions
Surjective functions
Bijective functions

3. Definition
We can symbolically define that a function is one-to-one (injective) with: f: XY is one-to-one  x1x2 f(x1) = f(x2)  x1 = x2

4. Definition
f: XY is one-to-one  x1x2 f(x1) = f(x2)  x1 = x2 From this statement, how can we determine the definition of a function that is not a one-to-one function? Take the negation. DO IT x1x2 f(x1) = f(x2)  x1  x2

5. Equivalent Definitions (contrapositive)
Mine: f: XY is one-to-one  x1x2 f(x1) = f(x2)  x1 = x2 Your book’s: x1x2 x1  x2  f(x1)  f(x2)

6. Activity 2 Here is a function. f: RR f(x) = 2x + 3
Is it a one-to-one function?
Could you prove it ?

7. Activity 2
To be one-to-one we must show that x1x2 f(x1) = f(x2)  x1 = x2 [definition of one-to-one function] Suppose x1 and x2 are real numbers such that f(x1) = f(x2) 2x1 + 3 = 2x2 + 3 substitution of definition of f(x) 2x1 = 2x2 Algebra x1 = x2 Algebra  f(x) is a one-to-one function

8. Activity 3 Here is a function. f: RR f(x) = x2 + 4
Is it a one-to-one function?
Could you prove it ?

9. Activity 3
Use a Counter example Let x1 = 2 and x2 = -2 f(x1) = = 8 f(x2) = (-2)2 + 4 = 8 Different values of x give same value of f(x)

10. Definition
We can symbolically define that a function is onto (surjective) with: f: XY is onto  yY xX f(x) = y

11. Definition
How could we define that a function is NOT onto (surjective) ? f: XY is not onto  yY xX f(x)  y

12. Activity 4 Return to the function. f: RR f(x) = 2x + 3
Is it an onto function?
Could you prove it ?

13. Activity 4
Suppose that y is a particular but arbitrary real number. We need to show that there is some real number x whose image is y. If such a real number exists, then 2x + 3 = y definition of our function x = (y - 3)/2 algebra x is a real number closure  f(x) = 2x + 3 is an onto function

14. Activity 5 Return to the function. f: RR f(x) = x2 + 4
Is it an onto function?
Could you prove it ?

15. Activity 5 We need to find a counterexample.
That is, we need to find a value in the target domain such that
yY xX f(x)  y
y = -1
No value of x plugged into x2 + 4 gives value of of -1