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?

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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?

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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 $\displaystyle (-\infty, 1]$

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