# Finding inverse when function conatains absolute value

1. Nov 17, 2013

### Persimmon

1. The problem statement, all variables and given/known data

Given the function f(x) = (abs(x))*x +6, find f^-1(x)

2. Relevant equations

3. The attempt at a solution

for x≥ 0, f(x) = x^2 + 6
y=x^2 +6
x = √(y-6) for y≥6
→ f^-1(x) = √(x-6) for x≥6

for x< 0, f(x) = -x^2 + 6
y= -x^2 +6
x = √(6-y) for y<6
→ f^-1(x) = √(6-x) for x<6

f^-1(x) = ±√(x-6) with the domain and range as all real numbers.
How do I get the inverse function to not be piecewise?
Sorry about the lack of proper equations, I don't know how to use latex.

2. Nov 17, 2013

### 462chevelle

have to restrict its not a 1 to 1

3. Nov 17, 2013

### Persimmon

Could you please elaborate? How is the function not one to one? f(x) = (abs(x))*x +6 results in a graph that essentially looks like a cubic function due to the absolute value sign.

4. Nov 17, 2013

### 462chevelle

alright. sorry, the way it was written caused me to jump the gun.
if you just solve it without the restrictions on x that's what you get.
but other than that im not sure. will have to wait on someone smarter.

5. Nov 17, 2013

### LCKurtz

You should have $x =\pm\sqrt{6-y}$ and you need to choose the - sign. So you would have
$f^{-1}(x) = -\sqrt{6-x},~x<6$.

Their "correct answer" is wrong. Your two piece answer is correct once you fix that minus sign.

6. Nov 17, 2013

### Persimmon

Yes, of course that makess sense regarding the minus sign. Oops. The reason why I specified that particular answer, ie ± √(x-6) is because that is what WolframAlpha gives as the result to the inverse of the original function. I can't think of how to arrive to that result.

7. Nov 17, 2013

### 462chevelle

8. Nov 17, 2013

### Persimmon

It's confusing me, because that's the book answer and the wolfram answer! There must be some arithmetic trick.

9. Nov 17, 2013

### LCKurtz

If you are confused, draw a graph of your original $f(x)$ and your $f^{-1}(x)$ on the same picture. If your inverse is the reflection of the original in the $45^\circ$ line, yours is correct. That's all there is to it.

10. Nov 17, 2013

### 462chevelle

11. Nov 17, 2013

### LCKurtz

What's interesting about that? Neither is a complete and correct picture of what I suggested in post #9. Can't you just do it by hand picking a few nice numbers?

12. Nov 17, 2013

### 462chevelle

I guess if youre just plugging numbers in for y and solving for x with both -(6-x)^1/2 and (x-6)^1/2
you get the same x value.
nevermind that. I keep leaving out stupid details.
why are wolfram and the book both wrong? is there something else to consider?

Last edited: Nov 17, 2013
13. Nov 17, 2013

### LCKurtz

They are wrong because they are wrong, what else is there to say. In post #2 you described the graph of $f(x)$. In post #5 I showed you a sign correction so you have $f^{-1}(x)$. As I said on post #9, graph them both on the same graph. Do it by hand if you can't make Wolfram behave with two piece functions.