range for e^|f(x)|

**Bushy** · March 25th 2010, 04:01 AM

Hi all, just looking at this function

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

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

These are correct.

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

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

**HallsofIvy** · March 25th 2010, 06:28 AM

Originally Posted by Bushy

Hi all, just looking at this function

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

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

These are correct.

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

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, $e^{|x+1|}$ is never less than 1, $-e^{x+1}$ is never larger than -1 and so $y= 2- e^{|x+1|}$ is never larger than 1. Of course when x= -1, |-x-1|= 0, $e^{-x-1}= 1$ and y= 1. Since |x+1| can be arbitrarily large and $e^x$ is an increasing function, y can be any negative number. That gives the range as $(-\infty, 1]$

**Bushy** · March 25th 2010, 03:35 PM

Thank you, this method seems obvious now it has been presented.

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