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Question
If f : R → (0, 2) defined by `f (x) =(e^x - e^(x))/(e^x +e^(-x))+1`is invertible , find f-1.
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Solution
Injectivity of f :
Let x and y be two elements of domain (R), such that
f (x) = f (y)
⇒ `(e^x - e^(-x))/(e^x -e^(-x)) +1 =(e^y - e^(-y))/(e^y -e^(-y)) + 1`
⇒`(e^x - e^(-x))/(e^x -e^(-x))= (e^y - e^(-y))/(e^y -e^(-y))`
⇒ `(e^-x(e^(2x) -1))/(e^-x(e^(2x)+1)) = (e^-y(e^(2y) -1))/(e^-y(e^(2y)+1)) `
⇒ `(e^(2x) -1)/(e^(2x) +1) = (e^(2y) -1)/(e^(2y) +1)`
⇒ (e2x−1) (e2y+1) = (e2x+1) (e2y−1)
⇒ e2x+2y + e2x−e2y −1= e2x+2y − e2x + e2y − 1
⇒ 2 × e2x =2 × e2y
⇒ e2x = e2y
⇒ 2x = 2y
⇒ x = y
So, f is one-one.
Surjectivity of f:
Let y be in the co-domain (0,2) such that f(x) = y.
`(e^x - e^-x)/(e^x +e^-x) + 1 = y `
⇒ `(e^-x(e^(2x) -1))/(e^-x(e^(2x)+1))+1 = y`
⇒ `(e^-x(e^(2x) -1))/(e^-x(e^(2x)+1)) = y - 1`
⇒ `e^(2x) -1 = (y - 1) (e^(2y) + 1)`
⇒ `e^(2x) -1 = y xx e^(2x) +y - e^(2x) -1`
⇒ `e^(2x) = y xx e^(2x) + y -e^(2x)`
⇒ `e^(2x) (2- y) = y`
⇒ `e^(2x) = y/(2-y)`
⇒ `2x = log_e (y/(2-y))`
⇒ `x = 1/2 log_e (y/(2 -y)) in R` (domain)
So, f is onto.
∴ f is a bijection and, hence, it is invertible.
Finding f -1:
Let f−1 (x) = y ...(1)
⇒ f (y) = x
⇒ `(e^y - e^-y)/(e^y + e^-y )+ 1 = x`
⇒ `(e^-y(e^(2y) -1))/(e^-y(e^(2y)+1)) + 1 = x`
⇒ `(e^-y(e^(2y) -1))/(e^-y(e^(2y)+1)) = x -1`
⇒ e2y −1 = ( x −1) ( e2y + 1 )
⇒ e2y − 1 = x × e2y + x − e2y − 1
⇒ e2y = x × e2y+ x − e2y
⇒ e2y ( 2 − x ) = x
⇒ `e^(2y) = x/(2-x)`
⇒`2y = log_e (x/(2-x))`
⇒`y =1/2 log_e (x/(2-x)) in R` (domain)
⇒`y =1/2 log_e (x/(2-x)) = f^-1 (x)` [from (1)]
` So, f^-1 (x) = 1/2 log_e (x/(2-x))`
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