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$f(x)=(x+2)^2$ for $x\geq-2$.

  1. Graph the function $f$.
  2. Find the slope of the tangent to the graph of $f^{-1}$ at the point with coordinates $(25,3)$ on the graph $f^{-1}$.

The answer according to a study guide gives me $1/10$ for the slope. How do I get that?

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    $\begingroup$ Could you clean this up a bit? I am not quite sure what you are asking. What I can tell you is that the slope of the tangent at any point on a graph is the derivative. So finding the derivative may give you some insight. $\endgroup$ – Chris Mar 6 '14 at 1:58
  • $\begingroup$ Chris, this is how the question is asked in my textbook. I don't know how to clean it up anymore and of course you have to find the derivative. Dza, I'm not sure. Every time I try to find the slope of the tangent of the graph it gives me a different answer. According to the textbook its 1/10, but I get a different answer. $\endgroup$ – santiagorook Mar 6 '14 at 16:25
  • $\begingroup$ Zakaria, I'm pretty sure that f^-1(x)=2(x+2)*(1) according to the chain rule $\endgroup$ – santiagorook Mar 6 '14 at 16:28
  • $\begingroup$ @santiagorook this comment must be in the answer below, no? And to answer you, according to the definition of reciproque function: $f^{-1}(f(x))=x$. Your $f^{-1}(x)$ doesn't match with the definition. $\endgroup$ – zagoku Mar 11 '14 at 12:07
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You have, for $x \geq 2$ : $f^{-1}(x)=\sqrt{x}-2$. The tangent line equation in $x_0$ is $y=(f^{-1})'(x_0)(x-x_0)+f(x_0)$ where $(f^{-1})'$ is the derivative of $f^{-1}$ (the slope is $(f^{-1})'(x_0)$).

Now where you're stuck?

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