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Question
Consider f : R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with `f^-1 (x) = (sqrt (x +6)-1)/3 .`
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Solution
Injectivity of f :
Let x and y be two elements of domain` (R^+)`, such that
f(x)=f(y)
⇒ 9x2+6x−5=9y2+ 6y − 5
⇒ 9x2+6x=9y2+6y
⇒ x = y (As, x, y ∈ `R^+`)
So, f is one-one.
Surjectivity of f:
Let y is in the co domain (Q) such that f(x) = y
⇒ 9x2 + 6x - 5 = y
⇒ 9x2 +6x = y + 5
⇒ 9x2 + 6x +1 = y +6 (Adding 1 on both sides )
⇒ (3x +1)2 = y + 6
⇒ `3x +1 = sqrt(y + 6)`
⇒ `3x = sqrt (y + 6) -1`
⇒ `x = (sqrt (y + 6)-1)/3 in R^+` (domain)
f is onto.
So, f is a bijection and hence, it is invertible.
Finding `f^-1`
Let f−1(x) = y ...(1)
⇒ x = f (y)
⇒ x = 9y2+ 6y − 5
⇒ x + 5 = 9y2+6y
⇒ x + 6= 9y2+ 6y + 1 (adding 1 on both sides)
⇒ x + 6 = ( 3y + 1 )2
⇒3y+1=`sqrt(x +6)`
⇒ `3y = sqrt (x +6) -1`
⇒ `y = (sqrt (x+6)-1)/3`
`So, f^-1 (x) (sqrt (x-6)-1)/3 ` [from (1)]
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