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#1 2017-12-20 03:21:43

Vedanti
Member
Registered: 2017-12-19
Posts: 5

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!

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#2 2017-12-20 09:11:05

bob bundy
Administrator
Registered: 2010-06-20
Posts: 8,371

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

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#3 2017-12-20 10:36:02

Vedanti
Member
Registered: 2017-12-19
Posts: 5

Re: Help Again Please...

Thanks Bob!

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#4 2017-12-31 09:06:36

zetafunc
Moderator
Registered: 2014-05-21
Posts: 2,196
Website

Re: Help Again Please...

If
is invertible, then
is bijective, and therefore injective. Therefore:

, so there are 3 points of intersection.

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