Inverse function

**Referos** · January 30th 2009, 01:14 PM

Hi, can someone please help me with a problem? I have to find the inverse function, but I have NO idea about how to do it. To find inverses, we always used the "make x the subject of the equation, then swap x for y" method, but this doesn't seem to work here.

"Find $f^{-1}(x)$ given that $f(x)=\frac{\arcsin{x}}{\ln{x}}$ "

Thanks.

**mollymcf2009** · January 30th 2009, 09:11 PM

Actually it does work!

For the function $f(x) = \frac{arcsin(x)}{\ln{x}}$

$y = \frac{arcsinx}{\ln{x}}$

$y\ln{x} = arcsin(x)$

$arcsin{x} =(y\ln{x})$

$y = sin (x \ln{x})$

Hope that helps you!

**Rincewind** · January 30th 2009, 09:42 PM

Originally Posted by mollymcf2009

$arcsin{x} =(y\ln{x})$

$y = sin (x \ln{x})$

How does this last step work. If you are just exchanging x for y then shouldn't you get

$y = sin (x \ln{y})$

?

**mollymcf2009** · January 30th 2009, 11:10 PM

I'm sorry I should not have put parentheses around that second part

The answer is y = sin(x) ln(x)

**Rincewind** · January 31st 2009, 12:58 AM

Originally Posted by mollymcf2009

I'm sorry I should not have put parentheses around that second part

The answer is y = sin(x) ln(x)

Sorry. I still don't see how you get that from

$\arcsin x = y \ln x$

even if you do exchange variables.

I can't see how to express an inverse function of

$y = \frac{\arcsin x}{\ln x}$

I thought about the following...

$y \ln x = \arcsin x$

$\ln x^y = \arcsin x$

$x = \sin\left(\ln x^y\right)$

$x = \frac{1}{2i}\left(x^y - x^{-y}\right)$

so basically x is the root of an equation

$x^y - 2ix - x^{-y} = 0$

which for a general $y$ has no explicit solution.

but maybe I'm missing something...

**mr fantastic** · January 31st 2009, 02:42 AM

Originally Posted by Rincewind

Sorry. I still don't see how you get that from

$\arcsin x = y \ln x$

even if you do exchange variables.

I can't see how to express an inverse function of

$y = \frac{\arcsin x}{\ln x}$

I thought about the following...

$y \ln x = \arcsin x$

$\ln x^y = \arcsin x$

$x = \sin\left(\ln x^y\right)$

$x = \frac{1}{2i}\left(x^y - x^{-y}\right)$

so basically x is the root of an equation

$x^y - 2ix - x^{-y} = 0$

which for a general $y$ has no explicit solution.

but maybe I'm missing something...

The thing you're missing, and in fact the thing that we're all missing, is the exact question. We only have the OP's version of it .......

Here is a scenario that could lead to the posted question:

Given that $f(x) = \frac{\arcsin x}{\ln x}$ find the value of $f^{-1}\left( - \frac{\pi}{6 \ln 2}\right)$ .

Or perhaps a scenario where the value of the derivative of the inverse function is asked for ....

**Referos** · January 31st 2009, 03:13 AM

I hate myself.
Since you guys said that there was something missing, I checked the question again. The exercise was written in an extremely small font size, and it turns out that it didn't ask for $f^{-1}(x)$ but for $f'(x)$ , which is perfectly solvable.

Sorry (and I will bring a magnifying glass the next time I try to solve a math problem)!

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