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Find ( f -1 )'( a )
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first off...by inspection \[f^{-1}(2)=0\]
Thats not an option
then use \[\left[f^{-1}\right]'(a)=\frac{1}{f'\left( f^{-1}(a) \right)}\]
what I pointed out was just the first step in the problem
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Ahh, see, these are the option I am given
ok...use the two posts of mine above and you should get the answer.
So if I did this right, 4/pi?
no
only 3 options to go :)
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lol, thats not what I'm going for
that won't help much come actual class time
\[f'(x)=2x+\sec^2(\pi x/2)\pi/2\] \[\frac{1}{f'(f^{-1}(2))}=\frac{1}{f'(0)}=\frac{1}{\pi/2}=\frac{2}{\pi}\]
Oh I see, you simplified f^-1 and then you got f'
just combined my first two posts in this thread.
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yeah, I'm kind of a slow/visual learner... but I'll get it
good
thanks man,
No problem
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