Help finding the inverse of a function?

g(x)=(x-2)^(1/2) 1

To find the inverse,
(1) Rename g(x) to y...........y = (x-2)^(1/2) 1
(2) Switch x and y...............x = (y-2)^(1/2) 1
(3) Solve for y.....................x - 1 = (y-2)^(1/2)
(x-1)² = (y-2)
(x-1)² 2 = y
y = (x-1)² 2
(4) Rename y to g^-1(x)
g^-1(x) = (x-1)² 2
This is your inverse function

To find the inverse, just switch x and y and then solve for y
y is the same as g(x)


so x = (y-2)^(1/2)+1

x-1 = (y-2)^(1/2)
(x-1)^2 = (y-2)
(x-1)^2+2 = y

Using old tricks that you have already learned:

g(x) = 1 + sqrt(x-2)
replace g(x) with y
(because it is easier to write and keep track, especially in Yahoo!Answers.

y = 1 + sqrt(x-2)
isolate the square root on one side
y - 1 = sqrt(x-2)
square both sides (this gets rid of the sqrt)
[y - 1]^2 = [sqrt(x-2)]^2
[y - 1]^2 = x - 2
move the 2 to the other side
[y - 1]^2 + 2 = x
flip it around (an equality works both ways)
x = [y - 1]^2 + 2

Now, swap the symbols
g^(-1)(x) = (x-1)^2 + 2

This is a "long" method, but it usually works very well.

g takes x, subtracts 2, takes square root, then adds 1.

so reversing that you get

g inverse takes x, subtracts 1, then squares, then adds 2.

Putting it into symbols:

g^-1(x) = (x-1)^2 + 2.