I understand that as x approaches (+) infinity, e^x = infinity.

I also understand that as x approaches (-) infinity, e^x = 0. The graph makes sense.

So, as x approaches (-) infinity, -e^x = infinity... Is this true?

I understand that as x approaches (+) infinity, e^x = infinity.

I also understand that as x approaches (-) infinity, e^x = 0. The graph makes sense.

So, as x approaches (-) infinity, -e^x = infinity... Is this true?

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No--because you have not changed the argument of the function at all (the exponent on top of the e), you are going to get 0 again. It's as if you multiplied both sides of the second equation by -1, giving you -1 * 0 = 0

Another way to look at it is to recall that if you multiply a function by -1 it is reflected about the x-axis. If you look at e^x reflected in such a way you can see that as you go towards -infinity you again get 0.

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## Comments

You could add that, as x approaches -infinity, e^x approaches 0

^{+ }(zero from the positive side of the y-axis) and -e^x approaches 0^{-}(zero from the negative side of the y-axis).More simply, ie from a graph, as x approaches negative infinity, e^x approaches zero from above, and -e^x approaches zero from below.

OK... thanks! That's what I suspected. I think my solutions guide is wrong...

What does your solutions guide say? e^(-x) would approach infinity as x approaches negative infinity.