Exponential Disequation

Don't know what you've graphed either Melody.

Anyway,

\(\displaystyle e^{x}-3e^{-x} + 2=(e^{2x}+2e^{x}-3)/e^{x}\)

\(\displaystyle =(e^{x}-1)(e^{x}+3)/e^{x}\),

and that will be greater than zero whenever

\(\displaystyle e^{x}-1 > 0\) .

That is, whenever x > 0.

e^x-3e^(-x)+2>0 {nl} how can i solve this disequation?

\(f(x)=e^x-3e^{-x}+2>0\)

\(x\geq 0,54933\)   (laut Wertetabelle MatheGrafix)

                                  !

I'd do it by graphing too but I do not know what asinus has graphed???

I use Desmos Graphing Calculator for my graaphing - it is an only calculator.

I graphed     

\(y=e^x-3e^{-x}+2\)

and this is what I got.

So I can see that this is positive when      x > 0.663    (approximately)

Check the Desmos graph : https://www.desmos.com/calculator/mlfl2icxyf

It appears that this will be true when  x >  0

Looks like asinus has missed the 2 from the end of f(x).

i.e. working with exp(x) - 3exp(-x) rather than exp(x) - 3exp(-x) + 2.