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Question
Show that f: [–1, 1] → R, given by f(x) = `x/(x + 2)` is one-one. Find the inverse of the function f: [–1, 1] → Range f.
(Hint: For y in Range f, y = `f(x) = x/(x + 2)` for some x in [–1, 1] i.e., `x = (2y)/(1 - y)`)
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Solution
f: [–1, 1] → R is given as `f(x) = x/(x + 2)`.
Let f(x) = f(y).
⇒ `x/(x + 2) = y/(y + 2)`
⇒ xy + 2x = xy + 2y
⇒ 2x = 2y
⇒ x = y
∴ f is a one-one function.
It is clear that f: [–1, 1] → Range f is onto.
∴ f: [–1, 1] → Range f is one-one and onto and therefore, the inverse of the function:
f: [–1, 1] → Range f exists.
Let g: Range f → [–1, 1] be the inverse of f.
Let y be an arbitrary element of range f.
Since f: [–1, 1] → Range f is onto, we have:
y = f(x) for same x ∈ [–1, 1]
⇒ `y = x/(x + 2)`
⇒ xy + 2y = x
⇒ x(1 – y) = 2y
⇒ `x = (2y)/(1 - y), y ≠ 1`
Now, let us define g: Range f → [–1, 1] as
`g(y) = (2y)/(1 - y), y ≠ 1`
Now, (gof)(x) = g(f(x))
= `g(x/(x + 2))`
= `(2(x/(x + 2)))/(1 - x/(x + 2))`
= `(2x)/(x + 2 - x)`
= `(2x)/2`
= x
(fog)(y) = f(g(y))
= `f((2y)/(1 - y))`
= `((2y)/(1 - y))/((2y)/(1 - y) + 2)`
= `(2y)/(2y + 2 - 2y)`
= `(2y)/2`
= y
∴ `gof = I_"[–1, 1]"` and `fog = I_"Range f"`
∴ `f^(-1) = g`
⇒ `f^(-1) (y) = (2y)/(1 - y), y ≠ 1`
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