# inverse functions

• Jul 24th 2011, 12:18 AM
Arron
inverse functions
When

f(x)=x^7+2x^3+x-1

I have found

f^-1(3)=1

But I need to prove this.

Can anyone help?
• Jul 24th 2011, 12:23 AM
FernandoRevilla
Re: inverse functions
$\displaystyle f(1)=3$ and $\displaystyle f$ is strictly increasing.
• Jul 24th 2011, 01:25 AM
Arron
Re: inverse functions
I need to find f^-1, or the inverse of f. Not sure how to do this?
• Jul 24th 2011, 02:38 AM
FernandoRevilla
Re: inverse functions
Quote:

Originally Posted by Arron
I need to find f^-1, or the inverse of f. Not sure how to do this?

By definition of $\displaystyle f^{-1}$ , $\displaystyle f^{-1}(\{y\})=\{x\in\mathbb{R}:f(x)=y\}$ . As $\displaystyle f(1)=3$ we can conclude that $\displaystyle 1\in f^{-1}(\{3\})$ . If we prove that $\displaystyle f$ is strictly increasing then, no $\displaystyle x\neq 1$ satisfies $\displaystyle f(x)=3$ so, $\displaystyle f^{-1}(\{3\})=\{1\}$ . That was your initial question. Now, if you want a closed formula $\displaystyle f^{-1}(x)$ for all $\displaystyle x\in\mathbb{R}$ , better forget it. :)
• Jul 24th 2011, 02:58 AM
mr fantastic
Re: inverse functions
Quote:

Originally Posted by Arron
I need to find f^-1, or the inverse of f.
[snip]

• Jul 24th 2011, 11:56 AM
chisigma
Re: inverse functions
Quote:

Originally Posted by Arron
I need to find f^-1, or the inverse of f. Not sure how to do this?

If finding $\displaystyle f^{-1}(*)$ is a 'categoric imperative' , then a method based on advanced tecniques of complex analysis will be supplied without demonstration...

If the function the inverse of which has to be found is...

$\displaystyle y= x^{7} + 2\ x^{3} + x -1$ (1)

... then first with the variable substitution $\displaystyle z=1+y$ we obtain...

$\displaystyle z=f(x)= x^{7} + 2\ x^{3} + x$ (2)

Now the inverse $\displaystyle x=f^{-1} (z)$ of a function for which is $\displaystyle f(0)=0$ can be written in series expansion as...

$\displaystyle x= \sum_{n=1}^{\infty} a_{n}\ z^{n}$ (3)

... where the $\displaystyle a_{n}$ are given by...

$\displaystyle a_{n}= \frac{1}{n!}\ \lim_{x \rightarrow 0} \frac{d^{n-1}}{d x^{n-1}} \{\frac{x}{f(x)}\}^{n}$ (4)

In our case of course is...

$\displaystyle \frac{x}{f(x)} = \frac{1}{x^{6}+ 2\ x^{2} +1}$ (5)

... and the x as function of y will be...

$\displaystyle x= \sum_{n=1}^{\infty} a_{n}\ (1+y)^{n}$ (6)

Details of computation of the $\displaystyle a_{n}$ are left to the [young I suppose...] starter of the thread...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jul 25th 2011, 12:58 PM
Arron
Re: inverse functions

One final thing, I am trying to determine the value of (f^-1)'(3)

Would this = 1/f'(3) = -1
• Jul 25th 2011, 02:36 PM
FernandoRevilla
Re: inverse functions
Quote:

Originally Posted by Arron
Would this = 1/f'(3) = -1

No, it is $\displaystyle (f^{-1})'(3)=\dfrac{1}{f'(f^{-1}(3))}=\dfrac{1}{f'(1)}=\ldots$
• Jul 27th 2011, 01:04 PM
Arron
Re: inverse functions
would that be 1/13.
• Jul 27th 2011, 01:49 PM
FernandoRevilla
Re: inverse functions
Quote:

Originally Posted by Arron
would that be 1/13.

Right.
• Jul 27th 2011, 03:25 PM
mr fantastic
Re: inverse functions
Quote:

Originally Posted by Arron
would that be 1/13.

Why do you need to even ask!? You have been told exactly what to do. Surely at this level of study you can differentiate a polynomial, substitute x = 1 and then take the reciprocal of the result!