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प्रश्न
Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with `f^(-1)(y) = ((sqrt(y +6) - 1)/3)`
उत्तर
f: R+ → [−5, ∞) is given as f(x) = 9x2 + 6x − 5.
Let y be an arbitrary element of [−5, ∞).
Let y = 9x2 + 6x − 5.
`=> y = (3x + 1)^2 - 1 - 5 = (3x + 1)^2 - 6`
`=> (3x + 1)^2 = y + 6`
`=> 3x + 1 = sqrt(y+6)` [as `y >= 5 => y + 6 > 0`]
`=> x = (sqrt(y + 6) - 1)/3`
∴f is onto, thereby range f = [−5, ∞).
Let us define g: [−5, ∞) → R+ as g(y) = `(sqrt(y+6) - 1)/3`
We now have:
`(gof) (x) = g(f(x)) = g(9x^2 + 6x - 5)`
`= g((3x + 1)^2 - 6)`
`= sqrt((3x + 1)^2 - 6 + 6 - 1)/3`
`= (3x + 1 - 1)/3 = x`
And `(fog)(y) = f(g(y)) = f((sqrt(y + 6)-1)/3)`
`= [3((sqrt(y + 6) - 1)/3)+1]^2 - 6`
`= (sqrt(y + 6))^2 - 6 = y + 6 - 6 = y`
`:. gof = I_R and fog = I_((-5, oo)`
Hence, f is invertible and the inverse of f is given by
`f^(-1) (y) = g(y)= (sqrt(y + 6) - 1)/3`
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