If then what is ?
It is undefined isn't it? since a particular object maps to two images?
Can someone confirm this? thanks
so:
So the inverse of |x| is |x|?? (I got that by solving for both intervals.)
I thought it might be undefined, since I was thinking of the reflection of the graph in y=x, where 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:
And we solve for :
And for x<0:
so we have the inverse of an absoulte value is:
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 since it is not bijective, but each 'piece' of the piecewise function is invertible induividually.
so -\infty ,0)\to(0,\infty ) " alt="f-\infty ,0)\to(0,\infty ) " /> defined by has 0,\infty )\to(-\infty ,0) " alt="f^{-1}0,\infty )\to(-\infty ,0) " /> defined by
and defined by has defined by
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?