Question

fine the inverse of the function f(x)=(x to the power 2) +6/(x-2) where x!=2

Answer 1

Answer:

f(x)=x^2+6/(x-2)

put x =2

f(2)=2^2+6/(2-2)

= 4+6/0

=4

Answer 2

Solution:

The given function is

f(x) = (x² + 6)/(x - 2), x ≠ 2

First we show that f is invertible.

1) Let x₁, x₂ be two distinct elements in the set of real numbers, taken as the domain of f.

f(x₁) = (x₁² + 6)/(x₁ - 2)

f(x₂) = (x₂² + 6)/(x₂ - 2)

So, f(x₁) ≠ f(x₂) when x₁ ≠ x₂

Also f(x₁) = f(x₂) when x₁ = x₂

Thus, f is injective.

2) Let y be an arbitrary element in the set of real numbers, taken as the co-domain of f.

f(x) = y

or, (x² + 6)/(x - 2) = y

or, x² + 6 = y (x - 2)

or, x² - yx + (6 + 2y) = 0

∴ x = {y ± √(y² - 24 - 8y)}/2

Since y is a real number, {y ± √(y² - 24 - 8y)}/2 is also a real number. Therefore y has pre-images {y ± √(y² - 24 - 8y)}/2 in the domain of f. Since y is taken as arbitrary, each element in the co-domain of f has a pre-image under f. Therefore f is surjective.

Since f is injective and surjective, f is a bijection, and hence invertible.

We have found that each element y in the co-domain of f has pre-images {y ± √(y² - 24 - 8y)}/2

So f⁻¹ is defined by

f⁻¹(y) = {y ± √(y² - 24 - 8y)}/2, y is a real number

or, equivalently

f⁻¹(x) = {x ± √(x² - 24 - 8x)}/2