Let f : N → Y : f(x) = 4x2+12x + 15, Y = Range(f) show that f is invertible and find f-1.
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Answer:
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f is invertible
<=> f is one-to-one (injective) and onto (surjective) (i.e. f is bijective)
Since Y is defined as the range of f, it is automatically onto (surjective). This means that for every value y∈Y there is an x such that y=f(x).
So to show that f is invertible, we just need to show that f is one-to-one (injective)...
f(a) = f(b) for a, b ∈ N
=> 4a² + 12a + 15 = 4b² + 12b + 15
=> a² + 3a = b² + 3b
=> a² - b² + 3a - 3b = 0
=> ( a - b ) ( a + b ) + 3 ( a - b ) = 0
=> ( a - b ) ( a + b + 3 ) = 0
=> either a = b OR a + b + 3 = 0.
Since a, b ∈ N, it is not possible for a + b + 3 = 0. Therefore a = b.
We've shown that if f(a) = f(b), then a = b. That is, we've shown that f is one-to-one. Therefore f is invertible.
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To get a formula, let y = f^{-1}(x). We want a formula for y in terms of x.
y = f^{-1}(x)
<=> x = f(y)
<=> x = 4y² + 12y + 15
<=> 4y² + 12y + (15-x) = 0
<=> y = ( -12 ± √( 12² - (4)(4)(15-x) ) ) / (2)(4)
= ( -3 ± √( 9 - (15-x) ) ) / 2
= ( -3 ± √( x - 6 ) ) / 2
Since y ∈ N, so y is a positive value, only the "+" sign here is possible, so we conclude that
f^{-1}(x) = ( -3 + √( x - 6 ) ) / 2