# Determine if these are functions

1. Nov 18, 2012

### nicnicman

Hello everyone,
I just want to make sure I'm doing these problems correctly. Here they are

Are the following functions?

1. F : Z→Z where F(x) = 4/7x + 1
Answer: Not a function. F(1) is not an integer.

2. G : R→R where G(x) = {2x + 2 if x ≥ 0, x - 3 if x ≤ 0
Answer: Not a function, because x is not well-defined. For G(0) there are two outputs.

3. h : R→R where h(x) = { x^3 if x > 3, 2x - 3 if x ≤ 3

Any suggestions are welcome.

Thanks.

2. Nov 18, 2012

### micromass

Staff Emeritus
That's correct.

3. Nov 18, 2012

### nicnicman

Thanks for the help!

4. Nov 18, 2012

### HallsofIvy

Staff Emeritus
Just a nitpick- in 1 F certainly IS a function, just not from Z to Z.

5. Nov 18, 2012

### nicnicman

Good point. Maybe I should specify. Would something like this work?
For Z→Z, F(x) = 4/7x + 1 is not a function.

6. Nov 18, 2012

### haruspex

For once, I disagree with Halls. F:Z→Z was part of the definition you were given, so the definition as a whole is not a valid definition of a function.

7. Nov 18, 2012

### micromass

Staff Emeritus
I agree with this. The domain and the codomain are an essential part of a function.

8. Nov 18, 2012

### Ray Vickson

In 1: be careful, use parentheses. It makes a difference whether you mean F(x) = (4/7)x + 1 or F(x) = 4/(7x) + 1. In the first case F(x) is an integer whenever x is an integer multiple of 7, but in the second case F(x) is never an integer for any nonzero integer value of x.

RGV

9. Nov 20, 2012

### nicnicman

Ray, since the domain and codomain are defined as Z→ Z wouldn't either case, F(x) = (4/7)x + 1 or F(x) = 4/(7x) + 1, still have to produce an integer for every integer x?

Also, the function should have been written as F : Z→ Z where F(x) = (4/7)x + 1.

Last edited: Nov 20, 2012