BET YOU CANT find the inverse! (1 Viewer)

[Eddyah]

y = e^x + 1/e^x


I really have no idea.... I hope someone out there does...

 

[jb_nc]

x = 1/(exp(y)+1/exp(y))

 

[watatank]

ok, well if you graph it (add graphs of e^x and e^-x together) its a kind of parabola shape. you will find that it only has an inverse when x is positive or x is negative (include zero). so thats the range for your inverse

y = e^x + 1/e^x

swap x and y and make y the subject to get the inverse

x = e^y + 1/e^y

x.e^y = (e^y)^2 + 1

(e^y)^2 - x.e^y + 1 = 0

let a = e^y

a^2 - ax + 1 = 0

quadratic formula to find a

sub e^y back in for a and take log(base e) of both sides and you have y as ur subject

just remember the range of the inverse is either solely negative, or solely positive, depending on what you took as your domain

sorry explanation is quite garbled...does that help?

 

[Eddyah]

ok, well if you graph it (add graphs of e^x and e^-x together) its a kind of parabola shape. you will find that it only has an inverse when x is positive or x is negative (include zero). so thats the range for your inverse

y = e^x + 1/e^x

swap x and y and make y the subject

x = e^y + 1/e^y

x.e^y = (e^y)^2 + 1

(e^y)^2 - x.e^y + 1 = 0

let a = e^y

a^2 - ax + 1 = 0

quadratic formula to find a

sub e^y back in for a and take log(base e) of both sides and you have y as ur subject

just remember the range of the inverse is either solely negative, or solely positive, depending on what you took as your domain

sorry explanation is quite garbled...does that help?

Dear watatank.
<script type="text/javascript"> vbmenu_register("postmenu_2913953", t</script>
You're a fucking champion!

Kind regards,
Eddy

 

[TheQueenOfLeon]

ok, well if you graph it (add graphs of e^x and e^-x together) its a kind of parabola shape. you will find that it only has an inverse when x is positive or x is negative (include zero). so thats the range for your inverse

y = e^x + 1/e^x

swap x and y and make y the subject to get the inverse

x = e^y + 1/e^y

x.e^y = (e^y)^2 + 1

(e^y)^2 - x.e^y + 1 = 0

let a = e^y

a^2 - ax + 1 = 0

quadratic formula to find a

sub e^y back in for a and take log(base e) of both sides and you have y as ur subject

just remember the range of the inverse is either solely negative, or solely positive, depending on what you took as your domain

sorry explanation is quite garbled...does that help?

Yes, yes, I see... ok, yep. keep going...

Dropping maths was the best decision of my life.

 

[watatank]

Yes, yes, I see... ok, yep. keep going...

Dropping maths was the best decision of my life.

yet you come into the maths forums..why?

and someone has posted the actual solution anyway

 

[TheQueenOfLeon]

Because maths people have a strange air of intelligence and I have a suppressed desire to be among them. (Actually, that's partially true)

I know it was.

 

[TheQueenOfLeon]

I am happy about what you said.

My imagination was females hate mathematicians and mathematics.

And next you could maybe imagine some utopian society in which all members of a sex don't adhere to a rigid standard of though?
I think the fact that there are people who can actually do maths is incredible. English is great but so easy to bullshit your way through, which I do to great success.

Our only 4U maths student is female.