The inverse of the function f(x)=ex−e−xex+e−x+2 is given by

let y=f(x) ⇒y-2=ex-e-xex+e-x apply compodendo divedendo y-1y-3=2ex-2e-x ⇒e2x=y-13-y ⇒x=12log(y-13-y) ⇒f-1(x)=log(x-13-x)12

The inverse of the function f(x)=ex−e−xex+e−x+2 is given by

let y=f(x) ⇒y-2=ex-e-xex+e-x apply compodendo divedendo y-1y-3=2ex-2e-x ⇒e2x=y-13-y ⇒x=12log(y-13-y) ⇒f-1(x)=log(x-13-x)12

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The inverse of the function $f (x) = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} + 2$ is given by

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\log_{e} {(\frac{x - 1}{x + 1})}^{- 2}$

$\log_{e} {(\frac{x - 2}{x - 1})}^{1 / 2}$

$\log_{e} {(\frac{x}{2 - x})}^{1 / 2}$

$\log_{e} {(\frac{x - 1}{3 - x})}^{1 / 2}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$l e t y = f (x) \Rightarrow y - 2 = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} a p p l y c o m p o d e n d o d i v e d e n d o \frac{y - 1}{y - 3} = \frac{2 e^{x}}{- 2 e^{- x}} \Rightarrow e^{2 x} = \frac{y - 1}{3 - y} \Rightarrow x = \frac{1}{2} \log (\frac{y - 1}{3 - y}) \Rightarrow f^{- 1} (x) = \log {(\frac{x - 1}{3 - x})}^{\frac{1}{2}}$