# Inverse function

1. Jan 27, 2005

### preet

How do I find the inverse of the following function: f(x)= -3x^2 + 6x + 2
The answer is +/- sqrt( [2-x/3] +1 ) + 1

I have no idea how to get that answer. I tried switching x and y and solving for x but I can't get the answer. TiA.

2. Jan 27, 2005

### vincentchan

use complete square.....

3. Jan 27, 2005

### preet

I get +/- sqrt(x-5/-3) +1

4. Jan 27, 2005

### dextercioby

Can u solve a quadratic equation???

If so,then solve for $x=x(y)$ from the equation
$$y=-3x^{2}+6x+2$$

Or you can complete the square...Correctly...

Daniel.

5. Jan 27, 2005

### preet

walk me through it... completing the square, I get -3(y-1)^2 + 5 ... and I wrote the inverse of that. I just want to know what I'm doing wrong...

6. Jan 27, 2005

### vincentchan

if your answer read like +/- sqrt((x-5)/-3) +1, then your answer is completely same as the textbook answer... why do you need help?

7. Jan 27, 2005

### dextercioby

you mean with "x" as in
$$y=-3(x-1)^{2}+5$$

Now solve for "x"...

Daniel.

8. Jan 27, 2005

### HallsofIvy

Staff Emeritus
Strictly speaking, f(x)= -3x2+ 6x+ 2 doesn't have an inverse!

Set y= -3x2+ 6x+ 2 and then use the quadratic formula to solve the quadratic equation -3x2+ 6x+ (2- y)= 0. You will see a + or - in the formula. In order to get an inverse, we need to restrict the domain. Since the vertex of the parabola is (1, 0.5), if x< 1 we get one inverse (with the -) and if x> 1 we get the other.

9. Jan 27, 2005

### preet

Yeah, I drew the graph... thats why I put +/- ... + is one inverse, - is the other

... I see that it's the same now... I dont get why they're different though... I mean I could tack on +1 -1 etc. How was the textbook answer found... (just out of curiosity)?

TiA