# Which one of these functions do NOT have an inverse?

• Sep 28th 2008, 12:51 PM
nugiboy
Which one of these functions do NOT have an inverse?
I know that it has to be the function that is 'many to one', but im having trouble finding that one.

Heres the four answers to choose from.

A) f(x) = 1 / (x - 1)

B) f(x) = x^3 + 1

C) f(x) = x^2 + 3 (where x is bigger/equal to 0)

D) f(x) = x^3 - 2x

I sort of worked out it it's going to be one to do with powers and negative numbers, which can be used to make the same product, but im can't seem to work out which one. I've only used trial and error really and a bit of logic. Can someone give me a hint or even better the answer and explanation?

..and sorry about my lack of math script skills.
• Sep 28th 2008, 01:42 PM
skeeter
find f(1) and f(-1) for answer (d)
• Sep 28th 2008, 01:49 PM
nugiboy
Quote:

Originally Posted by skeeter
find f(1) and f(-1) for answer (d)

Sorry but i tried that and it doesn't work.

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

Im pretty sure anyway.
Thanks though. Any other suggestions?
• Sep 28th 2008, 02:09 PM
11rdc11
Did you try graphing them and using the horizontal line test?
• Sep 28th 2008, 02:29 PM
nugiboy
Quote:

Originally Posted by 11rdc11
Did you try graphing them and using the horizontal line test?

Hey thanks for the idea. I should have thought of that. I just tried all the functions through on autograph, and ive hit a small problem. Graph D won't draw. All the other graphs seem to pass the test, so by elimination im guessing its graph D which won't. But i don't want to make assumptions. Would graph d pass the test? I can't really tell if it will or not.
• Sep 28th 2008, 02:33 PM
11rdc11
Nope D does not pass the test
• Sep 28th 2008, 02:48 PM
skeeter
Quote:

Originally Posted by nugiboy
Sorry but i tried that and it doesn't work.

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

Im pretty sure anyway.
Thanks though. Any other suggestions?

sorry ... was thinking \$\displaystyle f(x) = x^3 - x\$ rather than \$\displaystyle f(x) = x^3 - 2x\$

fact that it has more than one root is reason enough.