Indicate whether sech(x) is invertible for [0, infinity) and explain why

[yoohoo]

Homework Statement

Indicate whether sech(x) is invertible for [0, infinity) and explain why.

The Attempt at a Solution

Sooo... I know that:

sech(x) = 2 / ( e^x + e^-x )​

I know that to get the inverse equation I'd need to swap the y and the x... but I'm trying to show whether it's invertible so I don't know how much that would do me. I think I need to prove that the equation is monotonic, ie the derivative should be > 0 ). Hence, I took the derivative of the equation to be:

y' = - 2 (e^x - e^-x) / (e^x + e^-x)²​

Is that right? If so, is that enough to answer the question?

Thanks!

 

[SammyS]

You could show that sech(x) is not 1 to 1.

What are sech(1) and sech(-1) ?

 

[HallsofIvy]

But the problem said invertible on [0, infinity) so x= -1 is not relevant.

 

[SammyS]

DUH !

Sorry, I missed the [0, infinity) !

It's right there in the title and in the text!