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anonymous

  • 4 years ago

Find ( f -1 )'( a )

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  1. anonymous
    • 4 years ago
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  2. Zarkon
    • 4 years ago
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    first off...by inspection \[f^{-1}(2)=0\]

  3. anonymous
    • 4 years ago
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    Thats not an option

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  4. Zarkon
    • 4 years ago
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    then use \[\left[f^{-1}\right]'(a)=\frac{1}{f'\left( f^{-1}(a) \right)}\]

  5. Zarkon
    • 4 years ago
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    what I pointed out was just the first step in the problem

  6. anonymous
    • 4 years ago
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    Ahh, see, these are the option I am given

  7. Zarkon
    • 4 years ago
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    ok...use the two posts of mine above and you should get the answer.

  8. anonymous
    • 4 years ago
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    So if I did this right, 4/pi?

  9. Zarkon
    • 4 years ago
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    no

  10. Zarkon
    • 4 years ago
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    only 3 options to go :)

  11. anonymous
    • 4 years ago
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    lol, thats not what I'm going for

  12. anonymous
    • 4 years ago
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    that won't help much come actual class time

  13. Zarkon
    • 4 years ago
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    \[f'(x)=2x+\sec^2(\pi x/2)\pi/2\] \[\frac{1}{f'(f^{-1}(2))}=\frac{1}{f'(0)}=\frac{1}{\pi/2}=\frac{2}{\pi}\]

  14. anonymous
    • 4 years ago
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    Oh I see, you simplified f^-1 and then you got f'

  15. Zarkon
    • 4 years ago
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    just combined my first two posts in this thread.

  16. anonymous
    • 4 years ago
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    yeah, I'm kind of a slow/visual learner... but I'll get it

  17. Zarkon
    • 4 years ago
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    good

  18. anonymous
    • 4 years ago
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    thanks man,

  19. Zarkon
    • 4 years ago
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    No problem

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spraguer (Moderator)
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