If $\displaystyle f(x)=|x| $then what is $\displaystyle f^{-1}(x)$?
It is undefined isn't it? since a particular object maps to two images?
Can someone confirm this? thanks
so:
$\displaystyle 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 $\displaystyle f(f^{-1}(x))=x$ for both intervals.)
I thought it might be undefined, since I was thinking of the reflection of the graph in y=x, where $\displaystyle f:\{x\in\mathbb{R}:x>0\}\to\mathbb{R}$ 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.
I'm confused now. if:
$\displaystyle f(x)=|x|=\left\{\begin{array}{ll} x & \text{if x $\ge$ 0} \\ -x & \text{if x $<$ 0} \end{array} \right.
$
And we solve $\displaystyle f\circ f^{-1}(x)=x$ for $\displaystyle x\ge 0$:
$\displaystyle f\circ f^{-1}(x)=x \implies
f^{-1}(x)=x$
And for x<0:
$\displaystyle
f\circ f^{-1}(x)=x \implies
-f^{-1}(x)=x \implies f^{-1}(x)=-x$
so we have the inverse of an absoulte value is:
$\displaystyle
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.
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.However, both inverses are only defined for x > 0. The inverse of |x| is not |x|.
Both methods seem to be correct, but giving me different answers.
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
ah yes I see now The function |x| is not invertible over the domain $\displaystyle (-\infty , 0)\cup[0,\infty)$ since it is not bijective, but each 'piece' of the piecewise function is invertible induividually.
so $\displaystyle f-\infty ,0)\to(0,\infty ) $ defined by $\displaystyle f(x)= -x $ has $\displaystyle f^{-1}0,\infty )\to(-\infty ,0) $ defined by $\displaystyle f^{-1}(x)=-x$
and $\displaystyle f:[0,\infty )\to [0,\infty ) $ defined by $\displaystyle f(x)=x$ has $\displaystyle f^{-1}:[0,\infty )\to [0,\infty ) $ defined by $\displaystyle f^{-1}(x)=x$
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?