How can I find the right inverse of a function and show that a left one doesn't exist
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Let $f:mathbbR rightarrow [0,infty)$ be a mapping with $f(x)=x^2$ Show that $f$ has a right inverse, $h$, but not a left inverse and find h(0) and h(1)..
So from looking at this function, I know it's not injective because suppose $f(a) = f(b)$
So $a^2 = b^2$, then we have $pm a = pm b$. It's surjective because for $x = pm y$, $f(x) = y$
My idea of an inverse function would be:
let $h: [0,infty) rightarrow mathbbR$ be a mapping with $h(x) = x^frac12$
With this I can see that $f circ h = f(x^frac12) = x^(frac12)^2 = x$ so there is a right inverse but I can also see that
$h circ f = h(x^2) = (x^2)^frac12 = x$ which would mean it is a left inverse as well.
So what is wrong with my inverse function and how can I show that a right inverse exists but not a left one?
inverse-function
asked Sep 4 at 5:41
user8358234
815
add a comment |Â
up vote
0
down vote
favorite
Let $f:mathbbR rightarrow [0,infty)$ be a mapping with $f(x)=x^2$ Show that $f$ has a right inverse, $h$, but not a left inverse and find h(0) and h(1)..
So from looking at this function, I know it's not injective because suppose $f(a) = f(b)$
So $a^2 = b^2$, then we have $pm a = pm b$. It's surjective because for $x = pm y$, $f(x) = y$
My idea of an inverse function would be:
let $h: [0,infty) rightarrow mathbbR$ be a mapping with $h(x) = x^frac12$
With this I can see that $f circ h = f(x^frac12) = x^(frac12)^2 = x$ so there is a right inverse but I can also see that
$h circ f = h(x^2) = (x^2)^frac12 = x$ which would mean it is a left inverse as well.
So what is wrong with my inverse function and how can I show that a right inverse exists but not a left one?
inverse-function
asked Sep 4 at 5:41
user8358234
815
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f:mathbbR rightarrow [0,infty)$ be a mapping with $f(x)=x^2$ Show that $f$ has a right inverse, $h$, but not a left inverse and find h(0) and h(1)..
So from looking at this function, I know it's not injective because suppose $f(a) = f(b)$
So $a^2 = b^2$, then we have $pm a = pm b$. It's surjective because for $x = pm y$, $f(x) = y$
My idea of an inverse function would be:
let $h: [0,infty) rightarrow mathbbR$ be a mapping with $h(x) = x^frac12$
With this I can see that $f circ h = f(x^frac12) = x^(frac12)^2 = x$ so there is a right inverse but I can also see that
$h circ f = h(x^2) = (x^2)^frac12 = x$ which would mean it is a left inverse as well.
So what is wrong with my inverse function and how can I show that a right inverse exists but not a left one?
inverse-function
asked Sep 4 at 5:41
user8358234
815
Let $f:mathbbR rightarrow [0,infty)$ be a mapping with $f(x)=x^2$ Show that $f$ has a right inverse, $h$, but not a left inverse and find h(0) and h(1)..
So from looking at this function, I know it's not injective because suppose $f(a) = f(b)$
So $a^2 = b^2$, then we have $pm a = pm b$. It's surjective because for $x = pm y$, $f(x) = y$
My idea of an inverse function would be:
let $h: [0,infty) rightarrow mathbbR$ be a mapping with $h(x) = x^frac12$
With this I can see that $f circ h = f(x^frac12) = x^(frac12)^2 = x$ so there is a right inverse but I can also see that
$h circ f = h(x^2) = (x^2)^frac12 = x$ which would mean it is a left inverse as well.
So what is wrong with my inverse function and how can I show that a right inverse exists but not a left one?
inverse-function
inverse-function
asked Sep 4 at 5:41
user8358234
815
asked Sep 4 at 5:41
user8358234
815
asked Sep 4 at 5:41
user8358234
815
asked Sep 4 at 5:41
user8358234
815
asked Sep 4 at 5:41
user8358234
815
815
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
$h(f(x))=(x^2)^frac 1 2=x$ if $xgeq 0$ and $-x$ if $x<0$.
answered Sep 4 at 5:49
Kavi Rama Murthy
26k31437
but my domain is from $[0,infty)$ so why would I need to define it for $x<0$
– user8358234
Sep 4 at 5:51
Domain of $f$ is $mathbb R$ so you have to prove that $hcirc f$ is the identity map on $mathbb R$ if you want to claim that $h$ is the inverse of $f$.
– Kavi Rama Murthy
Sep 4 at 5:53
oh got it, thanks!
– user8358234
Sep 4 at 5:54
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$h(f(x))=(x^2)^frac 1 2=x$ if $xgeq 0$ and $-x$ if $x<0$.
answered Sep 4 at 5:49
Kavi Rama Murthy
26k31437
but my domain is from $[0,infty)$ so why would I need to define it for $x<0$
– user8358234
Sep 4 at 5:51
Domain of $f$ is $mathbb R$ so you have to prove that $hcirc f$ is the identity map on $mathbb R$ if you want to claim that $h$ is the inverse of $f$.
– Kavi Rama Murthy
Sep 4 at 5:53
oh got it, thanks!
– user8358234
Sep 4 at 5:54
add a comment |Â
up vote
2
down vote
accepted
$h(f(x))=(x^2)^frac 1 2=x$ if $xgeq 0$ and $-x$ if $x<0$.
answered Sep 4 at 5:49
Kavi Rama Murthy
26k31437
but my domain is from $[0,infty)$ so why would I need to define it for $x<0$
– user8358234
Sep 4 at 5:51
Domain of $f$ is $mathbb R$ so you have to prove that $hcirc f$ is the identity map on $mathbb R$ if you want to claim that $h$ is the inverse of $f$.
– Kavi Rama Murthy
Sep 4 at 5:53
oh got it, thanks!
– user8358234
Sep 4 at 5:54
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$h(f(x))=(x^2)^frac 1 2=x$ if $xgeq 0$ and $-x$ if $x<0$.
answered Sep 4 at 5:49
Kavi Rama Murthy
26k31437
$h(f(x))=(x^2)^frac 1 2=x$ if $xgeq 0$ and $-x$ if $x<0$.
answered Sep 4 at 5:49
Kavi Rama Murthy
26k31437
answered Sep 4 at 5:49
Kavi Rama Murthy
26k31437
answered Sep 4 at 5:49
Kavi Rama Murthy
26k31437
answered Sep 4 at 5:49
Kavi Rama Murthy
26k31437
26k31437
but my domain is from $[0,infty)$ so why would I need to define it for $x<0$
– user8358234
Sep 4 at 5:51
Domain of $f$ is $mathbb R$ so you have to prove that $hcirc f$ is the identity map on $mathbb R$ if you want to claim that $h$ is the inverse of $f$.
– Kavi Rama Murthy
Sep 4 at 5:53
oh got it, thanks!
– user8358234
Sep 4 at 5:54
add a comment |Â
but my domain is from $[0,infty)$ so why would I need to define it for $x<0$
– user8358234
Sep 4 at 5:51
Domain of $f$ is $mathbb R$ so you have to prove that $hcirc f$ is the identity map on $mathbb R$ if you want to claim that $h$ is the inverse of $f$.
– Kavi Rama Murthy
Sep 4 at 5:53
oh got it, thanks!
– user8358234
Sep 4 at 5:54
but my domain is from $[0,infty)$ so why would I need to define it for $x<0$
– user8358234
Sep 4 at 5:51
but my domain is from $[0,infty)$ so why would I need to define it for $x<0$
– user8358234
Sep 4 at 5:51
Domain of $f$ is $mathbb R$ so you have to prove that $hcirc f$ is the identity map on $mathbb R$ if you want to claim that $h$ is the inverse of $f$.
– Kavi Rama Murthy
Sep 4 at 5:53
Domain of $f$ is $mathbb R$ so you have to prove that $hcirc f$ is the identity map on $mathbb R$ if you want to claim that $h$ is the inverse of $f$.
– Kavi Rama Murthy
Sep 4 at 5:53
oh got it, thanks!
– user8358234
Sep 4 at 5:54
oh got it, thanks!
– user8358234
Sep 4 at 5:54
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2904667%2fhow-can-i-find-the-right-inverse-of-a-function-and-show-that-a-left-one-doesnt%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
