1 f: R -->R s.t f(x) = 2x +1
surjective. better to prove more explicitly:
let . We need to show s.t. . Note that . so let so that is onto.
Are the following functions injective, surjective or bijective
1 f: R -->R s.t f(x) = 2x +1
2 f: R -->R s.t f(x) = x^2
3 f: [0, oo) --> R s.t f(x) = x^2
4 f: [0, oo) -->[0, oo) s.t f(x) = x^2
5 f: R -->R s.t f(x) = e^x
6 f: R -->R s.t f(x) = log(x)
Whether a relationship is injective, surjective or bijective is really messing with my head and the standard definitons aren't helping me much. I'll try to explain what I think for each of these. These definitions help a little
Injective means that every member of "A" has a unique matching member in "B". You won't get two "A"s pointing to one "B", but you could have a "B" without a matching "A"
Surjective means that every "B" has at least one matching "A" (maybe more than one). Does this mean for every B being in the codomain?
1. Injective because every x has a unique corresponding f(x). Each x only maps to one f(x).
Surjective because for all of the codomain there is at least one x. Is this a correct way of thinking of it?
This means it is bijective.
2. Not injective because f(-2) = f(2). Not every x has a unique f(x).
Not surjective because the codomain is R and the function is not defined in R-.
3. Injective. By changing the domain to [0, oo) this removes the problem of each x (apart from zero) having two corresponding f(x).
Not surjective. Still not defined in R-.
4. Injective. Same as 3.
Surjective. By limiting the codomain to [0, oo) this sets the codomain equal to the range.
5. Injective. Each x maps to a unique f(x)
Not surjective. Parts of the codomain are undefined. Would this be surjective if we made the codomain R+
6. Not injective. R- of the domain is undefined for f(x).
Surjective. The whole codomain has been mapped onto by x.
Any comments on how I approached this? Mistakes? Thanks!