Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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 4 '13 at 11:31

1 Answer 1

up vote 3 down vote accepted

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.