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Question
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)`.
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Solution
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
⇒ y = (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, there by 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(9x2 + 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, ∞)`
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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