Thread: show that f(x) is injective but not surjective

Hi, firstly I've never really understood what injective and surjective means so if someone could give me the gist of that it'd be great!

secondly the question i'm stuck on at the moment is....

let A = {x:x (belongs to) [0, (inf))} and B = {x:x(belongs to) [0, inf)} and the function f: A (tends to) B is given by

f(x) = 1/(2+x^2)

show that f(x) is injective but not surjective

Thanks

Hello,

Try this:

Injectivity: http://en.wikipedia.org/wiki/Injective

Surjectivity: http://en.wikipedia.org/wiki/Surjective

thanks, a few minutes after posting i learned what they mean, im just unsure of how to prove it :s

Originally Posted by MrMathlete

Hi, firstly I've never really understood what injective and surjective means so if someone could give me the gist of that it'd be great!

secondly the question i'm stuck on at the moment is....

let A = {x:x (belongs to) [0, (inf))} and B = {x:x(belongs to) [0, inf)} and the function f: A (tends to) B is given by

f(x) = 1/(2+x^2)

show that f(x) is injective but not surjective

Thanks

Injective (one-to-one): If f(x)= f(y) then x= y. If show that x= y. It is important here that A consist of non-negative values only.

Surjective (onto): For any y in B the there exist x in A such that f(x)= y.
If , show that is not possible to solve for x for some values of y.