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P5 Hyperbolic Functions watch

1. Hiya, I've now moved on to studying P5 Hyperbolic functions and was hoping some of you guys can clear a couple of things up for me.

1) I know that tanh x is an odd function, but I'm confused with the proof the book provides:

tanh (-x) = [e^(-2x) - 1]/[e^(-2x) + 1]
= [1 - e^(2x)]/[1 + e^(2x)]
= - tanh x

Could someone explain how you go from the first step to the second step and why the e^(-2x) suddenly become e^(2x).

2) When drawing the graph of cothx, does it have assymptotes at y=+/- 1? Just my book doesn't show this graph and the assymptotes aren't shown on my graph drawing package (the one that does show assymptotes doesn't do hyperbolic functions! grrr).

Thanks ever so much

Chloé
2. tanh (-x) = [e(-2x)-1]/[e(-2x)+1)]

= [ (1/e^(2x)) - 1] / [ (1/e^(2x)) +1]

= [ (1-e^(2x)) / e^(2x)} ] / [ (1+e(2x)) / e^(2x)) ]

= [1-e(2x)] / [1+e(2x)]

= -tanh (x)

hope this helps....
3. (Original post by Hoofbeat)
Could someone explain how you go from the first step to the second step and why the e^(-2x) suddenly become e^(2x).
multiply top and bottom by e^(2x):

[e^(-2x).e^(2x) - 1.e^(2x)]/[e^(-2x).e^(2x) + 1.e^(2x)]

= [1 - e^(2x)]/[1 + e^(2x)]

(the one that does show assymptotes doesn't do hyperbolic functions! grrr)
then just draw y = [e^x + e^-x]/[e^x - e^-x]

but you can think about it....

as x -> v.large, e^-x -> 0, so cothx -> e^x/e^x = 1

and as x-> -v.large, e^x -> 0 => cothx -> e^-x/-e^-x = -1
4. Thanks ever so much you too, that helped loads!

Apologies for posting in wrong forum too! I swear I was in the Maths forum when I wrote message, but obviously not!

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Updated: August 24, 2004
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