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I have a question:

$\begin{array}{lrl} \mbox{If :} & f(x) & = x^5 + 3x^3 + 2x + 1 \\ \mbox{And :} & g(x) & = f^{-1} (x) \\ \mbox{What is :} & g'(7)&\mbox{?} \\ \mbox{What I do gives me :} &1/16 \end{array}$

But not sure how to proceed from here. Any help would be really appreciated. Thanks for checking my post $\require{enclose}\enclose{circle}{\ddot\smile}$

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One shortcut you can use is the fact that inverse functions have slopes that are reciprical at reflected points.

We want the slope ($g'(x)$ means slope) of $g(x)$ at $x=7$ or $(7,1)$. We know it is at (7,1) because $f(1)=7$ so the f(x) point we are reflecting is $(1,7)$ So you can find the slope of $f(x)$ at $x=1$ and take the reciprocal of that. Does that make sense?

EDIT: to clean that up a little bit...

  1. Find out where $f(x)=7$. A bit of solving shows that is at $x=1$.
  2. Find $f'(1)$
  3. $\frac{1}{f'(1)}=g'(7)$

I think this is correct. Haven't done inverse functions in awhile.

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  • $\begingroup$ I do not understand. If I cannot use the method you described, what might the regular flow look like? $\endgroup$
    – user40929
    Dec 11 '14 at 3:35
  • $\begingroup$ I can't get the inverse function outright, but I will edit my answer to show what i think is a valid method. $\endgroup$ Dec 11 '14 at 3:42
  • $\begingroup$ Would 1/16 be the answer? $\endgroup$
    – user40929
    Dec 11 '14 at 3:50
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    $\begingroup$ $\frac{1}{16}$ is what I got also. $\endgroup$ Dec 11 '14 at 3:51
  • $\begingroup$ Thanks! It was very helpful. $\endgroup$
    – user40929
    Dec 11 '14 at 4:03
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Simplifying on turkeyhundt's response, you can shorten the question to this:

\begin{equation} \text{What is } f^{-1}(7)\text{?} \end{equation}

Knowing that

\begin{equation} f^{-1}(y) = \frac{1}{f'(x)} \end{equation}

So find x, plug it into the equation, and you're good to go.

Solving for

\begin{equation} y = 7 = x^5+3x^3+2x+1 \end{equation}

You get

\begin{equation} x = 1 \end{equation}

Therefore,

\begin{equation} f^{-1}(7) = \frac{1}{f'(1)} \end{equation}

Giving \begin{equation}\frac{1}{16}\end{equation} as per the OP's original answer.

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  • $\begingroup$ Yes, much more succinct. On a couple of your lines where you have $f^{-1}(y)$, should it be the derivative of $f^{-1}(y)$? Like $f^{-1} $$'(y)$? $\endgroup$ Dec 11 '14 at 17:48
  • $\begingroup$ @turkeyhundt yeah actually it should be...thanks for pointing it out :) $\endgroup$ Dec 11 '14 at 23:58

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