# what's inverse of this function?

• Sep 1st 2008, 07:43 AM
what's inverse of this function?
CAN U TELL THE INVERSE OF
X-SIN(X)?
THANK U
• Sep 1st 2008, 07:50 AM
Soroban

First of all, release your CAPS LOCK . . .
All upper-case is very annoying ... and juvenile.

Quote:

Can you tell the inverse of: .$\displaystyle f(x) \:=\:x-\sin(x)$ ? . . . . no

We are required to solve for $\displaystyle y\!:\;\;y - \sin y \:=\:x$

This is a transcendental equaton . . . there is no elementary solution.

• Sep 6th 2008, 04:39 AM
I have recieved the answer as
$\displaystyle X+sin(X)$
Its mirror image
of $\displaystyle X-sin(X)$about $\displaystyle x=y$
• Sep 6th 2008, 09:04 AM
skeeter
Quote:

I have recieved the answer as
$\displaystyle X+sin(X)$
Its mirror image
of $\displaystyle X-sin(X)$about $\displaystyle x=y$

is that so?

http://i39.photobucket.com/albums/e1.../xplussinx.jpg
• Sep 6th 2008, 03:25 PM
Plato
Before I continue, let me say how dumb I think this question/answer is.
However, the posted answer is in fact correct.
If you graph the inverse correctly it is easy to see.
The reflection in the line y=x is $\displaystyle (x,y) \to (y,x)$.
• Sep 6th 2008, 04:10 PM
skeeter
are you saying that $\displaystyle y = x+\sin{x}$ is the inverse of $\displaystyle y = x-\sin{x}$ ?

here is a correct graph of $\displaystyle y = x-\sin{x}$ and its inverse $\displaystyle x = y-\sin{y}$ ...

http://i39.photobucket.com/albums/e1...hn/inverse.jpg
• Sep 7th 2008, 10:10 PM
I also got the same graph as Skeeter fr$\displaystyle x+sinx$
and that's why I posted this quest.
if reflection abt
$\displaystyle x=y$always the inverse
than wat abt
$\displaystyle sinx-x$
• Sep 8th 2008, 08:04 AM
Plato
Quote:

Originally Posted by skeeter
are you saying that $\displaystyle y = x+\sin{x}$ is the inverse of $\displaystyle y = x-\sin{x}$ ?

Absolutely not! Soroban has is correct. There is no elementary inverse possible.
However, in parametric form $\displaystyle \left( {t,t + \sin (t)} \right)\,\& \,\left( {t + \sin (t),t} \right)$ are reflections of each other in the line $\displaystyle y=x$ and as such act as an inverse.
In this new graphic, that point is illustrated using $\displaystyle t=2$.
• Sep 8th 2008, 04:09 PM
skeeter
my mistake ...

when you said

Quote:

However, the posted answer is in fact correct.
I thought you were talking about the OP's response to Soroban's response.