Question

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

Answer 1

Answer:

Step-by-step explanation:

Given is a function

$f(x) =(x-2)^2$

This function is a parabola with vertex at (2,0) and axis of symmetry is x=2

Hence for x<2 the curve would be reflection of x>2

To get inverse we must get one to one funciton only.

So restrict the domain of f(x) to $[2,∞)$

Then we have f(x) as one to one with domain x≥2 and range is R+

$f^{-1} (x)=+\sqrt{x} +2$

For this inverse domain is R+ and range is $[2,∞)$

Answer 2

Answer:

restriction of domain is x>=2

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

Step-by-step explanation:

Restrict the domain of the function  $f(x)=(x-2)^2$ so it has an inverse

To restrict the domain we find the vertex

VErtex form of the equation is

$y=(x-h)^2+k$ vertex is (h,k). restriction of domain is x>=h

$f(x)=(x-2)^2+0$ , vertex is (2,0)

So restriction of domain is x>=2

now we find inverse function

$f(x)=(x-2)^2$

Replace f(x) with y

$y=(x-2)^2$

Replace x with y and y with x

$x=(y-2)^2$

To remove square we take square root on both sides

$\sqrt{x} =y-2$

Add 2 on both sides

$\sqrt{x}+2 =y$

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