Finding inverse when function conatains absolute value

[Persimmon]

Homework Statement

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

Homework Equations

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

But the correct answer is
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.

 

[462chevelle]

have to restrict its not a 1 to 1

 

[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.

 

[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 I am not sure. will have to wait on someone smarter.

 

[LCKurtz]

Homework Statement

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

Homework Equations

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

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##.

But the correct answer is
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.

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

 

[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.

 

[462chevelle]

out of curiosity i checked wolfram and it has the same answer as the book answer. I wonder why there is a difference of opinion on this? maybe a minor detail is missing from the question?
http://www.wolframalpha.com/input/?i=inverse+|x|x+6

 

[Persimmon]

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

 

[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.

 

[462chevelle]

well. I plugged you guys answer in and interesting enough...
http://www.wolframalpha.com/input/?i=inverse+-((6-x)^1/2)
then there is this
http://www.wolframalpha.com/input/?i=inverse+-((x-6)^1/2)

 

[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?

 

[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?

 

[LCKurtz]

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?

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.