inverse functions (1 Viewer)

[jemsta]

does anyone know how to find the inverse of x=1/2(e^x-e^-x)?

 

[speed2]

do you mean y=1/2(e^x-e^-x)?

 

[speed2]

is the answer by any chance y=ln(x±√((x^2)-1))?? i'll show the working if iam right (small chance though)

 

[jemsta]

yeah speed is right

 

[speed2]

kool actually got 1 right. ok here is the method:
first swap around x and y so you get

x=1/2(e^y-e^-y)

then i let u = e^y, so you get

2x=u-1/u,

then multiply through by u so you get

u^2 - 2xu -1=0

then use the quadratic formula to find what u equals, so you'll get

u=(x±√((x^2)+1)), dat means that

e^y = (x±√((x^2)+1)),

then just log both sides.

EDIT: looked messy before so decided to clean it up a bit

 

[jemsta]

haha u absolute legend. thanks very much buddy
another 400 pages to go.............

 

[NT-social]

ok...man that's genius but i found something wrond...silly mistake
canu just check..isn't it in the ln(......+ 1) it should be plus one instead of minus

 

[jemsta]

yep
all fixed
not worth posting a simple mistake like that but thanks for pointiing it out anyways

 

[speed2]

ok...man that's genius but i found something wrond...silly mistake
canu just check..isn't it in the ln(......+ 1) it should be plus one instead of minus

yeah should be +1, damn silly mistakes!

 

[NT-social]

but seriously speed2 where and how did u come up with that method of solving...i dunno but i don't remember EVER solving it like that..just WOW!! >>skills

 

[jemsta]

haha....so if u ever find a question like this in the hsc exam....ull know which method to use.

 

[NT-social]

i reckon..it's seriously an eye-opener

 

[acmilan]

Out of interest: You'll find that (e^x - e^-x)/2 is a function called sinh (pronounced 'shine'). ie sinh x = (e^x - e^-x)/2

This function, along with its counterpart cosh x = (e^x + e^-x)/2 have remarkably similar properties to the circular functions ie. sin x and cos x, for example:

(cosh x)' = sinh x
(sinh x)' = cosh x
cosh²x - sinh²x = 1

cosh 2x = 2sinh²x + 1
cosh 2x = 2cosh²x - 1
cosh 2x = cosh²x + sinh²x

sinh 2x = 2 sinhx coshx

All of the above are easily proven with the definition of cosh and sinh

 

[speed2]

but seriously speed2 where and how did u come up with that method of solving...i dunno but i don't remember EVER solving it like that..just WOW!! >>skills

i've seen this type of question before, i remember the first time i looked at it i had no idea aswell.