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The problem I'm attempting to tackle says to take the derivative of the inverse, or ($f^{-1})'(a)$ of $\sqrt{x^3+x^2+x+22}$ when $a = 5$. From what I understand, I'm supposed to use the fact that ($f^{-1})'(a)$=$\frac{1}{f'(f^{-1}(a))}$. I was having a really hard time figuring out what $f^{-1}$ was, however, so I decided to punch it into Wolfram Alpha and see what it came up with for the inverse. I don't have high enough reputation to post an image, but here's the link for what I typed in. http://www.wolframalpha.com/input/?i=inverse+of+sqrt(x%5E3%2Bx%5E2%2Bx%2B22)

Am I doing something wrong here, or missing some important step? Did I do something wrong with Wolfram Alpha? Or is this genuinely the answer that I needed for $f^{-1}(x)$ and I'm just supposed to plug 5 into that and keep solving from there? I'm only in the first chapter of Calculus II, and that giant equation seems a bit crazy for being in one of the first assignments of the semester. Thanks for any clarification you can offer.

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    $\begingroup$ The point of that formula is that you do [b]not[/b] need to find $f^{-1}(x)$! $\endgroup$ – user247327 Jan 21 '17 at 20:07
  • $\begingroup$ But the formula requires $f^{-1}(x)$ in the denominator, does it not? $\endgroup$ – Potatomaster Jan 21 '17 at 20:20
  • $\begingroup$ Apparently I was not allowed to edit my above comment. You do need to find $f^{-1}(5)$ (not $f^{-1}(x)$) which means you need to solve $\sqrt{x^3+ x^2+ x+ 22}= 5$. Squaring both sides, $x^3+ x^2+ x+ 22= 25$ so you need to solve $x^2+ x^2+ x- 3= 0$. Happily, (you say that this was "in one of the first assignments of the semester") it is easy to see that x= 1 (the other two roots are complex). So $f'(5)= 1$ and you need to find $\frac{1}{f'(1)}$. $\endgroup$ – user247327 Jan 21 '17 at 20:23
  • $\begingroup$ You don't exactly need to find out what f^-1 (x) is $\endgroup$ – Saketh Malyala Jan 21 '17 at 20:29
  • $\begingroup$ @user Thank you so much, I was exceedingly frustrated trying to figure out what in the world I was doing wrong. I'll have to go back and read over this part in the textbook again since I obviously didn't understand it correctly the first time. :P $\endgroup$ – Potatomaster Jan 21 '17 at 20:36
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It is really more simple:

from $f(x)=\sqrt{x^3+x^2+x+22}$ finde the derivative: $$ f'(x)=\frac{3x^2+2x+1}{2\sqrt{x^3+x^2+x+22}} $$

now, as you say, use the formula: $$ (f^{-1})'(5)=\frac{1}{f'(f^{-1}(5))} $$ and note,by a simple inspection, that we have $f(x)=5$ for $x=1$ (and since the function is invertible, this is the only real value), so $f^{-1}(5)=1$.

Can you take it from here?

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