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Question
Let f : R `{- 4/3} `- 43 →">→ R be a function defined as f(x) = `(4x)/(3x +4)` . Show that f : R - `{-4/3}`→ Rang (f) is one-one and onto. Hence, find f -1.
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Solution
The function f : R - `{-4/3}→ R - {4/3}` is given by f (x) = `(4x)/(3x +4).` Injectivity: Let `x,y in R - {-4/3} `be such that f (x) = f (y)
⇒ `(4x)/(3x + 4)= (4x)/(3y+4)`
⇒ 4x (3y +4) = 4y (3x +4)
⇒ 12xy + 16x = 12xy +16y
⇒ 16x = 16y
⇒ x = y
Hence, f is one-one function.
Surjectivity: Let y be an arbitrary element of `R - {4/3}.` Then,
f(x) = y
⇒ `(4x)/(3x+4) = y`
⇒ 4x = 3xy +4y
⇒ 4x - 3xy = 4y
⇒ `x = (4y)/(4-3y)`
As `y ∈ R -{4/3} , (4y)/(4-3y) in R.`
Also , `(4y)/(4 - 3y) ≠ -4/3` because `(4y)/(4-3y) = - 4/3 ⇒ 12y = -16 +12y ⇒ 0 = -16,`which is not posssible.
Thus,
`x = (4y)/(4-3y) in R - {- 4/3}` sich that
`f (x) = f ((4x)/(3x +4))= (4((4y)/(4 -3y)))/(3((4y)/(4 -3y))+4) = (16y)/(12y+ 16 - 12y)= (16y)/16` = y , so every element in `R - {4/3} ` has pre-image in `R- {-4/3}.`
Hence, f is onto.
Now,
`x = (4y)/(4 -3y)`
Replacing x by f-1 and y by x, we have
`f^-1 (x) = (4x)/(4 - 3x)`
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