Math Is Fun Forum

Vedanti

Member

Registered: 2017-12-19

Posts: 7

Help Again Please...

I’m really stuck on this problem:

Suppose f(x) is a function defined for all real x, and suppose f is invertible (that is the inverse of f(x) exists for all x in the range of f).

If the graphs of y=f(x^2) and y=f(x^4) are drawn, at how many points do they intersect?

Thanks in advance!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,640

Re: Help Again Please...

hi Vedanti

Here again I wanted to try some graphs.  Here's a function plotter:

http://www.mathsisfun.com/data/function-grapher.php?

I tried f(x) = 2x + 1 first so I plotted 2x^2 + 1 and 2x^4 + 1

Then I tried f(x) = e^x  and then 1/x.  (strictly that one isn't invertible for all x but I ignored the one inadmissible value at this stage)

Definitely a common feature so I suggest you try it.  I'll hide my conclusion so you can try it for yourself first,

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Vedanti

Member

Registered: 2017-12-19

Posts: 7

Re: Help Again Please...

Thanks Bob!

zetafunc

Moderator

Registered: 2014-05-21

Posts: 2,440

Website

Re: Help Again Please...

If

is invertible, then

is bijective, and therefore injective. Therefore:

, so there are 3 points of intersection.