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Question
Let `f: R - {-4/3} → R` be a function defined as `f(x) = (4x)/(3x + 4)`. The inverse of f is map g: Range `f → R - {-4/3}` given by
Options
`g(y) = (3y)/(3 - 4y)`
`g(y) = (4y)/(4 - 3y)`
`g(y) = (4y)/(3 - 4y)`
`g(y) = (3y)/(4 - 3y)`
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Solution
`bb(g(y) = (4y)/(4 - 3y))`
Explanation:
It is given that `f: R -{-4/3} → R` is defined as `f(x) = (4x)/(3x - 4)`.
Let y be an arbitrary element of Range f.
Then, there exists `x ∈ R - {-4/3}` such that y = f(x).
⇒ `y = (4x)/(3x + 4)`
⇒ 3xy + 4y = 4x
⇒ x(4 – 3y) = 4y
⇒ `x = (4y)/(4 - 3y)`
Let us define g: Range `f → R -{-4/3}` as `g(y) = (4y)/(4 - 3y)`.
Now, (gof)(x) = g(f(x))
= `g((4x)/(3x + 4))`
= `(4((4x)/(3x + 4)))/(4 - 3((4x)/(3x + 4)))`
= `(16x)/(12x + 16 - 12x)`
= `(16x)/16`
= x
And (fog)(y) = f(g(y))
= `f((4y)/(4 - 3y))`
= `(4((4y)/(4 - 3y)))/(3((4y)/(4 - 3y)) + 4)`
= `(16y)/(12y + 16 - 12y)`
= `(16y)/16`
= y
∴ `gof = I_(R - {-4/3})` and `fog = I_"Range f"`
Thus, g is the inverse of f i.e., f−1 = g.
Hence, the inverse of f is the map g: Range `f → R - {-4/3}`, which is given by `g(y) = (4y)/(4 - 3y)`.
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