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AmTran_Bus

  • one year ago

Inverse problem?

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  1. AmTran_Bus
    • one year ago
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    I am willing to work it out, just need help.

  2. AmTran_Bus
    • one year ago
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    The inverse is y=x^2+4, right?

  3. geerky42
    • one year ago
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    What is "Theorem"?

  4. geerky42
    • one year ago
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    Correct

  5. AmTran_Bus
    • one year ago
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    Hum. Well, this is online, but let me reference a book.

  6. AmTran_Bus
    • one year ago
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    Maybe this?|dw:1433387173701:dw|

  7. geerky42
    • one year ago
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    Looks right, but seem useless, since we can just use g(x) to derivative with.

  8. geerky42
    • one year ago
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    We know \[g(x) =f^{-1}(x)= x^2+4\]

  9. AmTran_Bus
    • one year ago
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    Right

  10. geerky42
    • one year ago
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    well, you should be able to handle the rest.

  11. AmTran_Bus
    • one year ago
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    I mean, do you just plug it in?

  12. geerky42
    • one year ago
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    well, take derivative of g(x), then plug in

  13. AmTran_Bus
    • one year ago
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    2x, or 8...

  14. geerky42
    • one year ago
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    \(g(x) = x^2+4\) So \(g'(x) = 2x\) Now plug in \(x=4\); \[g'(4) = \boxed{8}\]

  15. AmTran_Bus
    • one year ago
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    Yay!

  16. AmTran_Bus
    • one year ago
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    Thank you

  17. geerky42
    • one year ago
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    No problem

  18. AmTran_Bus
    • one year ago
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    Hey @geerky42, would I do this the exact same way?

  19. AmTran_Bus
    • one year ago
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    Its inverse is y= 2+x^2+tan(pi x/2)

  20. geerky42
    • one year ago
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    That's exactly same function as original function.

  21. geerky42
    • one year ago
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    its inverse should have arctan

  22. AmTran_Bus
    • one year ago
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    My bad.

  23. AmTran_Bus
    • one year ago
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    Can you help me continue on then? I haven't really studied that in this context.

  24. geerky42
    • one year ago
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    Did you find inverse? You can swap x and y to each other, then isolate tan(...)

  25. geerky42
    • one year ago
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    from there, take arctan.

  26. AmTran_Bus
    • one year ago
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    Eek man, why is this one so hard?

  27. geerky42
    • one year ago
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    Gee didn't realize that finding inverse would be hard lol...

  28. geerky42
    • one year ago
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    @ganeshie8 Can you help?

  29. ganeshie8
    • one year ago
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    I don't seem to remember the inverse derivative formula... deriving it is easy we knw that \[\large f(f^{-1}(x)) = x\] differentiating both sides gives \[\large f'(f^{-1}(x)){f^{-1}}'(x)=1 \] \[\large {f^{-1}}'(x) = \dfrac{1}{ f'(f^{-1}(x))}\]

  30. ganeshie8
    • one year ago
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    plugin \(x=2\)

  31. geerky42
    • one year ago
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    Hmm. What is \(f(0)\)? @AmTran_Bus

  32. geerky42
    • one year ago
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    It seems that we need to use reasoning here.

  33. AmTran_Bus
    • one year ago
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    I think we are maybe overthinking. Look at the choices they give as answers.

  34. ganeshie8
    • one year ago
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    can you find f(0) ?

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