inverse functions

**Arron** · July 24th 2011, 12:18 AM

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?

**FernandoRevilla** · July 24th 2011, 12:23 AM

$f(1)=3$ and $f$ is strictly increasing.

**Arron** · July 24th 2011, 01:25 AM

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

**FernandoRevilla** · July 24th 2011, 02:38 AM

Originally Posted by Arron

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

By definition of $f^{-1}$ , $f^{-1}(\{y\})=\{x\in\mathbb{R}:f(x)=y\}$ . As $f(1)=3$ we can conclude that $1\in f^{-1}(\{3\})$ . If we prove that $f$ is strictly increasing then, no $x\neq 1$ satisfies $f(x)=3$ so, $f^{-1}(\{3\})=\{1\}$ . That was your initial question. Now, if you want a closed formula $f^{-1}(x)$ for all $x\in\mathbb{R}$ , better forget it.

**mr fantastic** · July 24th 2011, 02:58 AM

Originally Posted by Arron

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

Why? (Read previous reply).

**chisigma** · July 24th 2011, 11:56 AM

Originally Posted by Arron

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

If finding $f^{-1}(*)$ 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...

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

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

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

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

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

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

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

In our case of course is...

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

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

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

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

Kind regards

$\chi$ $\sigma$

**Arron** · July 25th 2011, 12:58 PM

Thanks for your help everyone.

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

Would this = 1/f'(3) = -1

**FernandoRevilla** · July 25th 2011, 02:36 PM

Originally Posted by Arron

Would this = 1/f'(3) = -1

No, it is $(f^{-1})'(3)=\dfrac{1}{f'(f^{-1}(3))}=\dfrac{1}{f'(1)}=\ldots$

**Arron** · July 27th 2011, 01:04 PM

would that be 1/13.

**FernandoRevilla** · July 27th 2011, 01:49 PM

Originally Posted by Arron

would that be 1/13.

Right.

**mr fantastic** · July 27th 2011, 03:25 PM

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!

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