SOLUTION: Inverse function of f(x) = (x+3)^2

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Question 1098187: Inverse function of f(x) = (x+3)^2
Found 2 solutions by josgarithmetic, Theo:
Answer by josgarithmetic(39797) About Me  (Show Source):
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
let y = f(x).

the equation becomes y = (x+3)^2.

replace y with x and x with y to get x = (y+3)^2

take the square root of both sides of the equation to get plus or minus sqrt(x) = y + 3

solve for y to get y = -3 plus or minus sqrt(x).

if they are inverse equations of each other, then the value of y in the original equation should be equal to the value of x in the inverse equation and the value of x in the original equation should be equal to the value of y in the inverse equation.

your original equation is y = (x+3)^2.

when x = 6, y = 9^2 = 81

when x = -6, y = (-3)^2 = 9

your coordinate pairs from the original equation are (6,81) and (-6,9)

your corresponding coordinate pairs in the inverse equations should be (81,6) and (9,-6).

your first inverse equation is y = -3 + sqrt(x).

when x = 81, y becomes equal to -3 + sqrt(81) which is equal to -3 + 9 which is equal to 6.

your coordinate pair is therefore (81,6) which is the inverse coordinate pair of (6,81) from the original equation.

your second inverse equation is y = -3 - sqrt(x).

when x = 9, y becomes equal to -3 - sqrt(9) which is equal to -3 - 3 which is equal to -6.

your coordinate pair is therefore (9,-6) which is the inverse coordinate pair of (-6,9).

you can see what's going on in the following graph:

$$$

the original equation is y = (x+3)^2.

the inverse equations are y = -3 + sqrt(x) and y = -3 - sqrt(x).

you can see that the inverse equation of y = -3 + sqrt(x) coresponds to the right side of the original equation of y = (x+3)^2.

you can also see that the inverse equation of y = -3 - sqrt(x) corresponds to the left side of the original equation of y = (x+3)^2.

note that the inverse equation of y = (x+3)^2 is actually x = (y+3)^2.

the rest is just solving for y in that equation.