Question

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.
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Answer 1

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