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Question
If f : R → (−1, 1) defined by `f (x) = (10^x- 10^-x)/(10^x + 10 ^-x)` is invertible, find f−1.
Solution
Injectivity of f:
Let x and y be two elements of domain (R), such that
f (x)=f (y)
`⇒ (10^x - 10^-x)/( 10^x - 10^-x ` = `(10^y - 10^-y)/( 10^y - 10^-y`
⇒ `(10^-x (10^(2x) - 1))/(10^-y (10^(2x)- 1)) = (10^-y (10^(2y) - 1))/(10^-y (10^(2y) - 1))`
⇒ `(10^(2x) - 1)/(10^2x +1)` = `(10^(2y) - 1)/(10^2y +1)`
⇒ (102x - 1 ) (102y +1) = (102x +1) (102y -1)
⇒ 102x + 2y + 102x - 102y - 1 = 102x +2y - 102x +102y - 1
⇒ 2 ×102x = 2 ×102y
⇒ 102x = 102y
⇒ 2x = 2y
⇒ x = y
So, f is one-one.
Surjectivity of f:
Let y is in the co domain (R), such that f(x) = y
⇒ `(10^x - 10^-x)/(10^x +10^-x) = y`
⇒ `(10^-x (10^(2x )-1))/(10^-x (10^(2x )+1)) =y`
⇒ `10^(2x) - 1 = y xx 10^(2x) +y`
⇒ `10^(2x) (1-y) = 1 +y`
⇒ `10^(2x) = (1+y)/(1 - y)`
⇒ `2x = log ((1+y)/(1-y))`
⇒`x = 1/2 log ((1+y)/(1-y)) in R` (domai
⇒ f is onto.
So, f is a bijection and hence, it is invertible.
Finding f -1:
⇒ Let f-1 (x) = y ......... (1)
⇒ f(y) = x
⇒ `(10^y - 10^-y)/( 10^y + 10^-y ) = x`
⇒ `(10^-y (10^(2y )-1))/(10^-y (10^(2y )+1)) = x`
⇒ `10^(2y) - 1 = x × 10^(2y) + x`
⇒ `10^(2y) = (1+x)/(1-x)`
⇒ `2y = log ((1+x)/(1-x))`
⇒ `y = 1/2 log ((1+x)/(1-x)) `
`So , f^-1 (x) = 1/2 log ((1+x)/(1-x))` [from (1)]
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