Question

Restrict the domain of the function f(x) = (x-2)^2 so it has an inverse. Then determine its inverse function.

Answer 1

Answer:

Look to the bold answer down

Step-by-step explanation:

* Lets explain how to restrict the domain of the quadratic function

- The quadratic function is two-to-one function

- The inverse of it is one-to-two which is not a function

- So we can not find the inverse of the quadratic function until restrict its

 domain

- We restrict the domain at the x-coordinate of the vertex of the function

∵ f(x) = (x - h)² + k is the standard form of the quadratic function, where

  (h , k) are the coordinates of its vertex

- To restrict the domain we put x > h for the right part of the parabola

  or x < h for the left part of the parabola

* Lets solve the problem

∵ f(x) = (x - 2)²

∵ f(x) = (x - h)² + k is the standard form of the quadratic function

∴ h = 2 and k = 0

∴ The vertex of the parabola is (2 , 0)

- We will restrict the domain at x = 2

∴ The domain of the function f(x) to have inverse is x > 2 or x < 2

* The restriction domain is x > 2 or x < 2

- To find the inverse of the function switch x and y and solve for the

  new y

∵ f(x) = (x - 2)²

∵ f(x) = y

∴ y = (x - 2)²

- Switch x and y

∴ x = (y - 2)²

- take square root for both sides

∴ ± √x = y - 2

- Add 2 for both sides

∴ ± √x + 2 = y

∴ $f^{-1}=\sqrt{x}+2=====OR=====f^{-1}=-\sqrt{x}+2$

* For the domain x > 2 of f(x) the inverse is $f^{-1}(x) = \sqrt{x}+2$

 For the domain x < 2 of f(x) the inverse is $f^{-1}(x)=-\sqrt{x}+2$