Which of the following functions f : {0, 1, 2, 3} ! {0, 1, . . , 7} are one-to-one?
1) f(x) = x2 mod 8
2) f(x) = x3 mod 8
3)f(x)=(x^3-8) mod 8

4)f(x)=(x^3+2x) mod 8

5) f(0) = 3, f(1) = 1, f(2) = 4, f(3) = 1

1. No.

Since f(1)=f(3) and then f isn't one-to-one.

2. No

Since f(0)=f(2) and then f isn't one-to-one.

3. No

Since f(0)=f(2) and then f isn't one-to-one.

4. Yes

Since , then f is one-to-one

5. Since f(1)=f(3) and then, f isn't one-to-one

step-by-step explanation:

i hate black

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x