# range for e^|f(x)|

• Mar 25th 2010, 04:01 AM
Bushy
range for e^|f(x)|
Hi all, just looking at this function

$\displaystyle y = -e^{|-x-1|}+2$

I have found the y-intercept to be $\displaystyle 2-e$ and x- intercepts to be $\displaystyle -1\pm\ln(2)$

These are correct.

I have also found a cusp type maximum at (-1,1) . Therefore $\displaystyle y \in (-\infty,1]$

This range was found using trial and error, is there a way to find this range other way for this type of function?
• Mar 25th 2010, 06:28 AM
HallsofIvy
Quote:

Originally Posted by Bushy
Hi all, just looking at this function

$\displaystyle y = -e^{|-x-1|}+2$

I have found the y-intercept to be $\displaystyle 2-e$ and x- intercepts to be $\displaystyle -1\pm\ln(2)$

These are correct.

I have also found a cusp type maximum at (-1,1) . Therefore $\displaystyle y \in (-\infty,1]$

This range was found using trial and error, is there a way to find this range other way for this type of function?

First, of course, |-x-1|= |x+ 1|. Since that is never negative, $\displaystyle e^{|x+1|}$ is never less than 1, $\displaystyle -e^{x+1}$ is never larger than -1 and so $\displaystyle y= 2- e^{|x+1|}$ is never larger than 1. Of course when x= -1, |-x-1|= 0, $\displaystyle e^{-x-1}= 1$ and y= 1. Since |x+1| can be arbitrarily large and $\displaystyle e^x$ is an increasing function, y can be any negative number. That gives the range as $\displaystyle (-\infty, 1]$
• Mar 25th 2010, 03:35 PM
Bushy
Thank you, this method seems obvious now it has been presented.