Question

Prove that the function f : N → N, defined by f(x) = x² + x + 1 is one-one but not onto. Find the inverse of f: N → S, where S is range of f.

Answer 1

The given function is
f : N → N
f(x) = x² + x + 1

Let x₁, x₂ 6N

So let f (x₁) = f (x₂)

`x_1^2 + x_1 + 1 = x_2^2 + x_2 + 1`

`x_1^2 - x_2^2 + x_1 - x_2 = 0`

(x₁ - x₂) (x₁ + x₂ + 1) = 0
∵  x₂ = x₁
or x₂  = - x₁ - 1
∵ x₁ ∈ N
∴ x₁ - 1 ∈ N

So x₂ ≠ -x₁ - 1

∵  f (x₂) = f (x₁)  only for x₁ = x₂

So f(x) is one -one function.

∵ f (x) = x² + x + 1

`"f" ("x") = ("x" + 1/2)^2 + 3/4`

Which is an increasing function.

f(1) = 3
∵ Range of f(x) will be {3, 7, .....} Which is a subset of N.

So it is an into function. i.e., f(x) is not an onto function.

let  y = x² + x + 1

x² + x + 1 - y = 0

`"x" = (-1± sqrt((1 - 4 )(1 - "y")))/(2)`

`"x" = (-1 ± sqrt(4"y" -3))/(2)`

So two possibilities are there for `f^-1 ("x")`

`"f"^-1 ("x") = (-1 + sqrt(4"x" -3))/(2), (-1 - sqrt(4"x" -3))/(2)` and we know `"f"^-1 (3)` = 1 because `"f"(1) = 3`

so `"f"^-1 ("x") = (-1 + sqrt(4"x" - 3))/(2)`