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AmTran_Bus

  • one year ago

Is this easy derivative question correct?

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  1. AmTran_Bus
    • one year ago
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  2. AmTran_Bus
    • one year ago
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    You don't have to walk me through, I know I need to be doing the work, but I would sure appreciate it if you would tell me if it is wrong so I can re do it.

  3. AmTran_Bus
    • one year ago
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    @welshfella

  4. AmTran_Bus
    • one year ago
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    Anyone have an idea? I can post the formula

  5. AmTran_Bus
    • one year ago
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    Gotta love Patrick JMT on youtube

  6. zepdrix
    • one year ago
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    um um um

  7. AmTran_Bus
    • one year ago
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    lol yes zepdrix? :)

  8. zepdrix
    • one year ago
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    \[\Large\rm f(x)=2+x^2+\tan\left(\frac{\pi}{2}x\right)\]So we want to evaluate this at the inverse functions x=2, which will correspond to the original functions y=2, ya? \[\Large\rm 2=2+x^2+\tan\left(\frac{\pi}{2}x\right)\]So uhhh.. it looks like this is leading to an x value of x=0.

  9. zepdrix
    • one year ago
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    So then uhhhhh.. ya our formula stuff.

  10. AmTran_Bus
    • one year ago
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    Maybe so, looks good?

  11. zepdrix
    • one year ago
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    Ooo ya I made a boo boo somewhere :P Cause f(0) is = 2. So we would end up with 1/2. Woops. Not an option. Thinkingggg +_+

  12. AmTran_Bus
    • one year ago
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    Its alright I understand this problem is a doozy

  13. zepdrix
    • one year ago
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    Oh oh oh, I'm being silly.. it's not f(0) in the bottom it's f'(0)

  14. zepdrix
    • one year ago
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    ya ya ya, that should fix things up :) find your derivative before plugging 0 in

  15. zepdrix
    • one year ago
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    \[\Large\rm (f^{-1})(2)=\frac{1}{f'(\color{orangered}{f^{-1}(2)})}\] \[\Large\rm (f^{-1})(2)=\frac{1}{f'(\color{orangered}{0})}\]

  16. AmTran_Bus
    • one year ago
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    Eek whats the d dx?

  17. zepdrix
    • one year ago
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    Umm I guess it would be \[\Large\rm f(x)=0+2x+\sec^2\left(\frac{\pi}{2}x\right)\cdot\frac{\pi}{2}\]I applied chain rule on that last term.

  18. AmTran_Bus
    • one year ago
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    Ok so Im right then?

  19. zepdrix
    • one year ago
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    oh with the blue dot? ahhh yes, looks correct!! :)

  20. AmTran_Bus
    • one year ago
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    THANK YOU!!!! WHooo hoooo!!!!

  21. zepdrix
    • one year ago
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    yay team \c:/

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