# Math Help - Inverse of absolute value

1. ## Inverse of absolute value

If $f(x)=|x|$then what is $f^{-1}(x)$?

It is undefined isn't it? since a particular object maps to two images?

Can someone confirm this? thanks

2. Hello,
Originally Posted by Greengoblin
If $f(x)=|x|$then what is $f^{-1}(x)$?

It is undefined isn't it? since a particular object maps to two images?

Can someone confirm this? thanks
Define f as following :

$f(x)=\left\{\begin{array}{ll} x \quad \text{if x \ge 0} \\ -x \quad \text{if x < 0} \end{array} \right.$

Then define an inverse function for x > 0 and for x < 0. It should be okay from here

3. so:

$f^{-1}(x)= \left\{\begin{array}{ll} x & \text{if x \ge 0} \\ -x & \text{if x < 0}\end{array} \right.$

So the inverse of |x| is |x|?? (I got that by solving $f(f^{-1}(x))=x$ for both intervals.)

I thought it might be undefined, since I was thinking of the reflection of the graph in y=x, where $f:\{x\in\mathbb{R}:x>0\}\to\mathbb{R}$ is undefined due to any x having multiple value of f.

Cheers

P.S. Why isn't my code working for the first bit? It's almost exactly the same as your array.

4. Technically speaking, the inverse of |x| is not a function. However, the inverse of y = |x| for x>0 is y = x, and the inverse of y = |x| for x < 0 is y = -x. However, both inverses are only defined for x > 0. The inverse of |x| is not |x|.

5. Originally Posted by icemanfan
Technically speaking, the inverse of |x| is not a function. However, the inverse of y = |x| for x>0 is y = x, and the inverse of y = |x| for x < 0 is y = -x.
I'm confused now. if:

$f(x)=|x|=\left\{\begin{array}{ll} x & \text{if x \ge 0} \\ -x & \text{if x < 0} \end{array} \right.
$

And we solve $f\circ f^{-1}(x)=x$ for $x\ge 0$:

$f\circ f^{-1}(x)=x \implies
f^{-1}(x)=x$

And for x<0:

$
f\circ f^{-1}(x)=x \implies
-f^{-1}(x)=x \implies f^{-1}(x)=-x$

so we have the inverse of an absoulte value is:

$
f^{-1}(x)= \left\{\begin{array}{ll} x & \text{if x \ge 0} \\ -x & \text{if x < 0} \end{array} \right. = |x|$

So this is my reasoning for thinking the inverse of |x|=|x|. I can't see anything wrong with the way I've done this.

However, both inverses are only defined for x > 0. The inverse of |x| is not |x|.
When I think about the inverse graphically, it's not the graph of a function at all. Look at the attached image |x| in red, and it's inverse in blue (reflection in y=x) which is undefined for x<0 since it doesn't take x values in this range, and also for x>0, since any x here, maps to multiple values of y.

Both methods seem to be correct, but giving me different answers.

6. Actually, only bijective functions have an inverse one

f has inverse <=> f is bijective

f is bijective <=> f is injective and surjective

Now, if we define:
f:R ->R
f(x)=|x|
it has no inverse, since its not even injective

but if we define:
f:R+ ->R+
f(x)=|x|
we obtain a bijective function

so it depends on how you define the domain and the codomain

7. ah yes I see now The function |x| is not invertible over the domain $(-\infty , 0)\cup[0,\infty)$ since it is not bijective, but each 'piece' of the piecewise function is invertible induividually.

so $f-\infty ,0)\to(0,\infty ) " alt="f-\infty ,0)\to(0,\infty ) " /> defined by $f(x)= -x$ has $f^{-1}0,\infty )\to(-\infty ,0) " alt="f^{-1}0,\infty )\to(-\infty ,0) " /> defined by $f^{-1}(x)=-x$

and $f:[0,\infty )\to [0,\infty )$ defined by $f(x)=x$ has $f^{-1}:[0,\infty )\to [0,\infty )$ defined by $f^{-1}(x)=x$

So unless the domian is specified as otherwise, the function f=|x| is not invertible.

Thanks

ugh I always have problems with LaTeX. Anyone care to explain what is wrong this time?

8. right