# show that f(x) is injective but not surjective

• May 6th 2010, 05:53 AM
MrMathlete
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 (Happy)
• May 6th 2010, 08:35 AM
surjective
Injectivity - surjectivity
• May 6th 2010, 10:45 AM
MrMathlete
thanks, a few minutes after posting i learned what they mean, im just unsure of how to prove it :s
• May 7th 2010, 02:27 AM
HallsofIvy
Quote:

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 (Happy)

Injective (one-to-one): If f(x)= f(y) then x= y. If $\frac{1}{2+ x^2}= \frac{1}{2+ y^2}$ 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 $\frac{1}{2+ x^2}= y$, show that is not possible to solve for x for some values of y.