Discrete mathematics question (1 Viewer)

[HeroicPandas]

Hi, can someone please check my working out?

With g : R -> R defined by g(x) = x³ - x + 1

a) What is the range of f?
b) Is f onto? Explain.
c) Is f one-to-one? Explain.

My working: let E mean 'is an element of'

a) (-∞, ∞)

b)
Method 1: Draw x³ - x and shift it 1 unit up. Looking at the graph, g(x) is onto because for all y E R, there is at least one x E R such that f(x) = y

Method 2: g(x) is onto because the range of f, (-∞, ∞) is equal to the codomain of f (which is all real, which is (-∞, ∞))

c) No. Counter-example: g(0) = g(-1) = g(1) = 1.

If this question was 3 marks, what would I get? (what would I get if I used method 1, and what would I get if I used method 2)

Thanks

 

[integral95]

I do not know the marking criteria as i'm a first year noob.

But I would say:

For method 2, you should add that all of the elements in the codomain has at least one corresponding value in the domain.

 

[HeroicPandas]

I do not know the marking criteria as i'm a first year noob.

But I would say:

For method 2, you should add that all of the elements in the codomain has at least one corresponding value in the domain.

Yes yes

 

[Shadowdude]

Hi, can someone please check my working out?

With g : R -> R defined by g(x) = x³ - x + 1

a) What is the range of f?
b) Is f onto? Explain.
c) Is f one-to-one? Explain.

My working: let E mean 'is an element of'

a) (-∞, ∞)

b)
Method 1: Draw x³ - x and shift it 1 unit up. Looking at the graph, g(x) is onto because for all y E R, there is at least one x E R such that f(x) = y

Method 2: g(x) is onto because the range of f, (-∞, ∞) is equal to the codomain of f (which is all real, which is (-∞, ∞))

c) No. Counter-example: g(0) = g(-1) = g(1) = 1.

If this question was 3 marks, what would I get? (what would I get if I used method 1, and what would I get if I used method 2)

Thanks

If you drew a picture and used that as proof, I think you would get zero automatically.

 

[Carrotsticks]

Hi, can someone please check my working out?

With g : R -> R defined by g(x) = x³ - x + 1

a) What is the range of f?
b) Is f onto? Explain.
c) Is f one-to-one? Explain.

My working: let E mean 'is an element of'

a) (-∞, ∞)Correct.

b)
Method 1: Draw x³ - x and shift it 1 unit up. Looking at the graph, g(x) is onto because for all y E R, there is at least one x E R such that f(x) = yCorrect.

Method 2: g(x) is onto because the range of f, (-∞, ∞) is equal to the codomain of f (which is all real, which is (-∞, ∞)) Correct. I prefer this method as it does not require the sketch. Technically Method 1 is correct but it doesn't always work because you may not have a 'sketchable function'.

c) No. Counter-example: g(0) = g(-1) = g(1) = 1. Correct.

If this question was 3 marks, what would I get? (what would I get if I used method 1, and what would I get if I used method 2)

Thanks

Replied in bold.

 

[HeroicPandas]

If you drew a picture and used that as proof, I think you would get zero automatically.

OK, I'll need to ask my tutor about that

Replied in bold.

Thanks, very helpful