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

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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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,399

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

Thanks Bob!

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If is invertible, then is bijective, and therefore injective. Therefore:, so there are 3 points of intersection.

**LearnMathsFree: Videos on various topics.New: Integration Problem | Adding FractionsPopular: Continued Fractions | Metric Spaces | Duality**

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