Inverse of x+1/x

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Homework Statement



Find a section of the graph that is one-to-one and find the inverse of the graph for that domain.

The Attempt at a Solution


Well I've set the domain to all real numbers greater than 0. This makes the function one to one right?

Now how do you find the inverse of the function? I know you isolate for x then switch x and y but I can't seem to isolate for x...

I get stuck at

xy-x2=1

I'm pretty sure the inverse should be

1/2(x±√x2-4) Based on the quadratic equation, but how do I account for my domain restriction? This inverse function is not one-to-one...
 
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Answers and Replies

  • #2
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Ok so I'm pretty sure I need to restrict my domain in some other area, but how do I determine which area to restrict the domain?

Then how do I know which inverse function to choose?
 
  • #3
Hootenanny
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Ok so I'm pretty sure I need to restrict my domain in some other area, but how do I determine which area to restrict the domain?

Then how do I know which inverse function to choose?
You need to restrict your domain such that [itex]f(x) = x + 1/x[/itex] and
1/2(x±√x2-4)
are one-to-one.

Sketching the curves is always a good idea.
 
  • #4
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You need to restrict your domain such that [itex]f(x) = x + 1/x[/itex] and

are one-to-one.

Sketching the curves is always a good idea.
ok I've restricting the domain to [1, ∞), this is a one to one function of x+1/x right? It passes the horizontal line test.

I still don't know what to do about the inverse though, which do I choose? Is it even the correct inverse for this function? Every time I try going back and forth from regular to inverse the value is off by a small amount.

Edit: I've just re-read what you wrote, sorry. So both the inverse AND the original function must be one-to-one? Does such a domain even exist for this function?

Edit again: I've just graphed the inverse on wolfram (pretty awestruck it can just get the inverse like a search engine actually) and this is what it looks like when graphed WITH the original function:
http://www.wolframalpha.com/input/?i=inverse(x+1/x)

I don't think there's any location possible where both functions are one-to-one. Does this imply that a unique inverse function is not possible? Unless we can restrict the range as well as the domain? Or perhaps restricting the range on the original function restricts the domain on the second function limiting which function I should choose?
 
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  • #5
Hootenanny
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ok I've restricting the domain to [1, ∞), this is a one to one function of x+1/x right? It passes the horizontal line test.

I still don't know what to do about the inverse though, which do I choose? Is it even the correct inverse for this function? Every time I try going back and forth from regular to inverse the value is off by a small amount.

Edit: I've just re-read what you wrote, sorry. So both the inverse AND the original function must be one-to-one? Does such a domain even exist for this function?
Yes, you have the correct inverse. Here is a sketch of the function and its inverse:
[PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP61519hda0d8h042ab27000048b09576ce4663e6?MSPStoreType=image/gif&s=21&w=496&h=362 [Broken]
Decide for yourself whether the inverse is one-to-one anywhere. :wink:
 
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  • #6
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Yes, you have the correct inverse. Here is a sketch of the function and its inverse:
[PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP61519hda0d8h042ab27000048b09576ce4663e6?MSPStoreType=image/gif&s=21&w=496&h=362 [Broken]
Decide for yourself whether the inverse is one-to-one anywhere. :wink:
I'm going to hazard a guess that the inverse function of x+ 1/x is not unique so not "real" (is that the right term?) Seeing as the domain values for the inverse of the function for any given domain restriction will output two range values.
 
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  • #7
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Nevermind, I've figured it out. I was inputting the numbers wrong. Thanks for the help :)
 

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