Hello everybody.Can anyone give me an example of a function f(x) which is surjective but not injective.?

Question

f(x) is defined :

f:R->R.

I was thinking something like f(x)=xsinx for all x E R.Would this work?I think it would.I know its not injective but i cant yet show exactly why its surjective.Help appreciated.

Answer 1

One example would be

f(x) = x^3 - x.

This is surjective ("onto") but not injective ("one to one") because

f(0)=f(1)=f(-1) = 0

Another would be

f(x)= tan x.

EDIT: Your example also works. Note that the amplitude of the oscillations of Asinx is A. So here, as x gets large, the amplitude of the oscillations increase without bound (i.e., goes to + and - infinity) so it will be onto. (You should graph this to see it)

Here is a rigorous proof: consider x>0 and divide into intervals that go from x=pi/2+2pi*n to 3pi/2+2pi*n, n=0,1,2.... Since sin(pi/2+2pi*n)=+1 and sin(3pi/2+2pi*n)=-1, we have that at the left endpoint of each interval f(x)= pi/2+2pi*n and at the right it

equals  -(3pi/2+2pi*n). Now we need to show that given any real number c, there is some x=a such that f(a)=c. Any c must belong to some interval  

pi/2+2pi*n > c > -(3pi/2+2pi*n) for some choice of n (Archimedean property). But the function is continuous on a closed, bounded interval so the Intermediate Value Thm applies. Thus there must be some x=a in

[pi/2+2pi*n, 3pi/2+2pi*n] such that f(a)=c. So f is onto (surjective).