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TimmyG

  • one year ago

I need help with finding the inverse of this function, 2x^5+x^3+1.

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  1. TimmyG
    • one year ago
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    \[2x^5+x^3+1\]

  2. jdoe0001
    • one year ago
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    hmm, technically, the inverse is just swapping about the variables, now the simplification, well, not so simplistic

  3. TimmyG
    • one year ago
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    exactly, I know the concept it's just that I have to show my work to simplify and insert a value into the new function.

  4. klimenkov
    • one year ago
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    Interesting... Can somebody find the inverse function for \(y=x^2+2x-3\) ?

  5. pgpilot326
    • one year ago
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    yes... but you will have to restrict the domain.

  6. klimenkov
    • one year ago
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    Two branches will satisfy: \[-1\pm\sqrt{4+x}.\]Now how to solve this one? \[2y^5+y^3+1-x=0.\]

  7. TimmyG
    • one year ago
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    throwing it into wolframalpha gives this result http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427ef49hilhcgk

  8. klimenkov
    • one year ago
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    This means that the inverse function may not be found.

  9. TimmyG
    • one year ago
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    whaaaaaat...... But It's in my textbook and part of my homework! Ugh...

  10. pgpilot326
    • one year ago
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    what are you studying?

  11. TimmyG
    • one year ago
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    calc one... this is the second day of homework.

  12. pgpilot326
    • one year ago
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    you sure it's a 5th power?

  13. TimmyG
    • one year ago
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    sorry, that's the class. I'm studying Electrical Engineering. just a second and I'll upload a snapshot of the problem

  14. TimmyG
    • one year ago
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    #74

  15. klimenkov
    • one year ago
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    Hahaha. Lol.

  16. jdoe0001
    • one year ago
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    hmm well well

  17. pgpilot326
    • one year ago
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    dude... it says f^-1(a) and gives you the value of a. inverse functions work like this... plug in the y from a function into its inverse and you get out x. compute f(a) for your functions and then f^-1(f(a)) will just be a.

  18. TimmyG
    • one year ago
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    I know, and I have the -71 as the answer, but it also asks to write down the function which I can't find to save my life!

  19. pgpilot326
    • one year ago
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    oops... i misread. but i doesn't ask for the inverse function... only f^-1(a)

  20. TimmyG
    • one year ago
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    OH! oh oh oh oh.... Damn....

  21. pgpilot326
    • one year ago
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    if you find a = f(x) then the x that satisfies this will be f^-1(a)

  22. DebbieG
    • one year ago
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    Wow that's a lot easier than finding \(f^{-1}(x)\) :)

  23. TimmyG
    • one year ago
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    yeah, it's -71...

  24. TimmyG
    • one year ago
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    craaaaap.... well there went an hour I'll never get back. thanks for all your help guys

  25. pgpilot326
    • one year ago
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    no worries... maybe post the question from your book and your interpretation so you can get to the heart of the sunrise quicker.

  26. klimenkov
    • one year ago
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    So, what is an answer?

  27. pgpilot326
    • one year ago
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    it's not -71

  28. TimmyG
    • one year ago
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    no?

  29. pgpilot326
    • one year ago
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    x = -1 -2 = 2x^5 + x^3 +1 => 2x^5 + x^3 +3 = 0 the only real solution is x = -1

  30. TimmyG
    • one year ago
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    yeah I get what you mean now... but isn't that just f(-2) now? not the inverse?

  31. pgpilot326
    • one year ago
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    no... f(-1) = -2 so f^-1(-2) = -1

  32. DebbieG
    • one year ago
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    Remember that \(f^{-1}(-2)\) is that domain element x that gives you \(f(x)=-2\) The function maps the domain to the range. The inverse function maps the range back to the domain.

  33. klimenkov
    • one year ago
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    Let \(f^{-1}(a)=x\). Now map \(f\) to both of the parts of the equation and use \(f(f^{-1}(a))=a\). It will become\[a=f(x)\]Solve this and you will obtain an answer.

  34. DebbieG
    • one year ago
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    So by setting \(f(x)=y=-2\) and solving, you find the x in the domain that takes you to y in the range.

  35. pgpilot326
    • one year ago
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    f^-1(f(x)) = x

  36. TimmyG
    • one year ago
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    ohhh okay, so put the variable as the solution and work backwards

  37. pgpilot326
    • one year ago
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    they give you a = f(x) and want you to find f^-1(a) = f^-1(f(x)) = x so setting f(x) = a and solving for x gives you what you need.

  38. pgpilot326
    • one year ago
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    make sense?

  39. TimmyG
    • one year ago
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    Yep :) thanks again y'all

  40. pgpilot326
    • one year ago
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    you're welcome!

  41. TimmyG
    • one year ago
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    *lightbulb turns on* I think the rest of this will be much easier now

  42. pgpilot326
    • one year ago
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    awesome! let's turn the dark into light!

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