Let f:N → R be a function defined as f(x) = 4x² + 12x + 15. Show that f:N>S, where S is the range of f is invertible. Find the inverse of f.
Answers
Given : f : N → R be a function defined as f(x) = 4x² + 12x + 15.
To show : f : N → S , where S is range of f, is invertible.
To find : inverse of f.
solution : we know any function is invertible only when function is one - one as well as onto.
let two points x₁ and x₂ from its domain such that f(x₁) = f(x₂)
⇒4x₁² + 12x₁ + 15 = 4x₂² + 12x₂ + 15
⇒4(x₁² - x₂²) + 12(x₁ - x₂) = 0
⇒(x₁ - x₂)(x₁ + x₂) + 3(x₁ - x₂) = 0
⇒(x₁ - x₂)(x₁ + x₂ + 3) = 0.
as domain of f belongs to N [ all positive integers ]
so, (x₁ + x₂ + 3) ≠ 0. hence (x₁ - x₂) = 0
⇒x₁ = x₂
here f(x₁) = f(x₂) ⇒x₁ = x₂
Therefore f is one - one function.
y = f(x) = 4x² + 12x + 15
⇒4x² + 12x + 15 - y = 0
⇒x = {-12 ± √{12² - 4(4)(15 - y)}}/2(4)
= {-3 ± √(y - 6)}/2
but domain of f = x ∈ N
so, x can't be negative. i.e., x ≠ {-3 - √(y - 6)}/2
hence x = {-3 + √(y - 6)}/2
now putting value of x in f(x)
f(x) = 4[{-3 + √(y - 6)}/2]² + 12[{-3 + √(y - 6)}/2] + 15
= {-3 + √(y - 6)}² - 18 + 6√(y - 6) + 15
= (-3)² + {√(y - 6)}² - 6√(y - 6) - 18 + 6√(y - 6) + 15
= 9 + y - 6 - 3
= y
i.e., f(x) = y Thus, for every y in range of f, there is a pre - image x in N.
so co - domain = Range
Therefore f must be onto function.
here f is one - one and onto so f is invertible function.
and inverse of f = x = {-3 + √(y - 6)}/2
i.e., f¯¹(x) = {-3 + √(x - 6)}/2