# Inverse Functions

1. Feb 12, 2009

### jeahomgrajan

what is the inverse of f(x)=-2/3(x+5)^2 - 5/3

so far i got x=-2/3(y+5)^2 -5/3
bring 5/3 over to x side
x+5/3=-2/3(y+5)^2

although i know what to do, can someone just explain how they will go on from here

2. Feb 12, 2009

### affirmative

What you want to be doing, I think, is to substitute f(x)=y. Then, you need to switch x and y in the equation and rearrange it to get y=something. Did I explain myself clearly?

3. Feb 12, 2009

### jeahomgrajan

it is stated that i have begun doing that, so now i know i will have to divide -2/3 by x+5/3, then i will sqareroot the left side, as the right side is being squared, and the finally bring over the 5 which will reslut in -5

overall equation = (sqareroot)(x+5/3) - 5 =y
........................................ -2/3

(x+5/3 is being divided by -2/3)

4. Feb 12, 2009

### jeahomgrajan

i believe i have to get ride of the -2/3 how will i do so?
( i am trying to graph it so i should be able to apply the transformations)

5. Feb 12, 2009

### affirmative

I can only help you find the inverse in terms of y=something man. Anything other than that and I'm sorry -- have to wait for someone else to help you... :(

6. Feb 12, 2009

### Staff: Mentor

So far, so good. Now multiply both sides by -3/2, which gives you
-3/2 * (x + 5/3) = (y + 5)^2

If you want, you can switch sides to get

(y + 5)^2 = -3/2 * (x + 5/3)

Take the square root of both sides (but don't forget that you'll have +/-).
Then subtract 5 and you'll have y = f^(-1)(x)

7. Feb 13, 2009

### HallsofIvy

Strictly speaking, that doesn't have an inverse because of the square. As Mark44 said, when you take the square root, you will have a "$\pm$" so you do not have a well defined function.