# Homework Help: Some problems with me

1. Nov 19, 2006

### Broken_Mirage

Hi all,,

i hope you could help me in the two problem below:uhh:

1-2piCOS(2piX)=0

and

y=exp(x-1)-x
i want here x in terms of y

thanx for help

2. Nov 19, 2006

### HallsofIvy

??? It's not clear to me what you are asking. There is no "y" in the first equation so x cannot be written "as a function of y", it is a constant: $cos(2\pix)= 1/(2\pi)$ so x= 1/(2\pi)cos^{-1}(1/(2\
pi)[/itex].

If the second equation is independent of the first, the x can be written as a y, but not using "elementary" functions. You should be able to write it in terms of "Lambert's W function" which is defined as the inverse function to f(x)= xex.

3. Nov 21, 2006

### Broken_Mirage

hi man and thank you for reply,,

its big problem for me and im really want help:
the first equation is
f(x)=x-sin(2pix) and they want from me to find the absolute maximum and absolute minimum of f(x)????

the second equation is independent of the first
y=exp(x-1)-x

the question is
f(x)=exp(x-1)-x
find the inverse
so i want x=0.5y+1(for example)
i sure that you know the inverse, so i wait your answer

thank you for reply my dear

4. Nov 21, 2006

### lotrgreengrapes7926

There is no absolute max or min. $$\displaystyle\lim_{x\to\infty} x-\sin(2\pi x)=\infty$$ and $$\displaystyle\lim_{x\to-\infty} x-\sin(2\pi x)=-\infty$$.