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I know it is a simple problem but I am having trouble. Here is what I have so far:

Let $f(x) = x^5 + 2x^3 + x - 1$

a) Find $f(1)$ and $f'(1)$

I have a) done. $f(1)$ is $3$ and $f'(1)$ is $12$

b) Find $f^{-1}(3)$ and $(f^{-1})'(3)$

I need help with the first part. I think the way to find the inverse is to switch the $x$'s with $y$'s and then solve for $y$. But I am having trouble completing this. I have the following: $$ x = y^5 + 2y^3 + y -1 $$ $$ x - 1 = y^5 + 2y^3 + y $$ What am I supposed to do here?

Toadise
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    Not that it helps in the slightest, but it wouldn't be $x - 1 = y^5 + 2y^3 + y$. It would be $x + 1 = y^5 + 2y^3 + y$. But this doesn't get you anywhere. yunone is right, you don't have to find the inverse explicitly. – labyrinth Apr 04 '13 at 11:31

1 Answers1

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I don't think you need to find $f^{-1}$ explicitly. Remember that $f^{-1}(3)$ in this case is the number $x$ such that $f(x)=3$, and use what you've found in part (a).

For $(f^{-1})' (3)$, remember that by the inverse function theorem, $$ (f^{-1})'(b)=\frac{1}{f'(a)} $$ where $f(a)=b$, and again use what you've found in part (a).

yunone
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