# inverse function

• Oct 30th 2012, 07:12 PM
chizmin10
inverse function
let f^(-1) : R->R be the inverse function of f(x)= x^5+2x+2. Find (f^(-1))'(5)

1. So im guessing i take the inverse function of x^5+2x+2? but i'm having trouble figuring it out.

2. I think i take the derivative of the inverse function

3. And then i think i plug in 5

any hoo i need help to figure this out please and thank you
• Oct 30th 2012, 07:57 PM
TheEmptySet
Re: inverse function
Quote:

Originally Posted by chizmin10
let f^(-1) : R->R be the inverse function of f(x)= x^5+2x+2. Find (f^(-1))'(5)

1. So im guessing i take the inverse function of x^5+2x+2? but i'm having trouble figuring it out.

2. I think i take the derivative of the inverse function

3. And then i think i plug in 5

any hoo i need help to figure this out please and thank you

You will want to use the inverse function theorem.

If we take the derivative we get $f'(x)=5x^4+2$

Since this derivative is non zero the function is 1-1 and we can use the inverse function theorem.

To do so we need to solve the equation

$5=x^5+2x+2 \iff x^5+2x-3=0$

By the rational roots theorem the only possible zeros are $\pm 1, \pm 3$

Checking $x=1 \implies 1^5+2(1)-3=0$

Now by the inverse function theorem we have

$\frac{d}{dx}f^{-1}(x)\bigg|_{x=5}=\frac{1}{f'(1)}=\frac{1}{5(1)^4+ 2}=\frac{1}{7}$
By the rational roots theorem we
• Oct 30th 2012, 08:25 PM
chizmin10
Re: inverse function
Thank you very much, one last question why is 3 a possible zero?
• Oct 31st 2012, 03:50 AM
TheEmptySet
Re: inverse function
Quote:

Originally Posted by chizmin10
Thank you very much, one last question why is 3 a possible zero?

For any polynomial $p(x)=a_nx^n+a_{n-1}x^{n-1}+....+a_1x+a_0$

The possible rational zeros are $\frac{p}{q}$ where p and q are all the possible factors are $a_n$ and $a_0$