Inverse function question (1 Viewer)

[rama_v]

Hey guys

Quick question from the 2002 HSC ext maths paper (7.a.iii)

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2002exams/pdf_doc/mathemat_ext1_02.pdf

find an expression for y = f^-1(x) in terms of x assuming that f(x) = e^x + 1/(e^x)

Thanks

 

[shafqat]

Hey guys

Quick question from the 2002 HSC ext maths paper (7.a.iii)

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2002exams/pdf_doc/mathemat_ext1_02.pdf

find an expression for y = f^-1(x) in terms of x assuming that f(x) = e^x + 1/(e^-x)

Thanks

Firstly it should be 1/e^x
Secondly, after switching x and y, let u = e^y, and then solve the quadratic in u.

 

[rama_v]

Firstly it should be 1/e^x
Secondly, after switching x and y, let u = e^y, and then solve the quadratic in u.

Oops, typo fixed. Thanks, i got it now

 

[nick1048]

what did u end up getting for ur answer?

I get a rather ugly looking quadratic with logs

 

[KFunk]

Sounds about right. If you just switch the x's and y's then:

x = e^y + 1/e^y

(e^y)² - xe^y + 1 = 0 (quadratic in e^y)

e^y = (x ± √[x² - 4])/2

y = log_e((x - √[x² - 4])/2)

EDIT: Just now I had a look at it in graphmatica so if you got something like that then it should be correct.

 

[rama_v]

Yeah, thats what I got as well