How do I check if a function has an inverse function?
$begingroup$
From what I've learnt, a function $f$ has an inverse function $f^{-1}$ if the function $f$ is injective (one-to-one, horizontal rule applies).
How can I check if a function has an inverse in the first place? Given two examples:
1:
$f(x) = frac{arcsin x-1}{arcsin x+2}$
its inverse is:
$f^{-1}(x)= sin frac{-2x-1}{x-1}$
2:
$g(x) = frac{ln x}{x}$
its inverse is:
$g^{-1}(x)= e^{frac{-x}{x-1}}$
But how do I properly check if a function is an injective fuction? Preferably without using derivatives.
Thanks.
real-analysis analysis inverse-function
asked Dec 5 '18 at 2:46
wenoweno
1499
$endgroup$
add a comment |
$begingroup$
From what I've learnt, a function $f$ has an inverse function $f^{-1}$ if the function $f$ is injective (one-to-one, horizontal rule applies).
How can I check if a function has an inverse in the first place? Given two examples:
1:
$f(x) = frac{arcsin x-1}{arcsin x+2}$
its inverse is:
$f^{-1}(x)= sin frac{-2x-1}{x-1}$
2:
$g(x) = frac{ln x}{x}$
its inverse is:
$g^{-1}(x)= e^{frac{-x}{x-1}}$
But how do I properly check if a function is an injective fuction? Preferably without using derivatives.
Thanks.
real-analysis analysis inverse-function
asked Dec 5 '18 at 2:46
wenoweno
1499
$endgroup$
1
$begingroup$
Depending on context, I just want to note: whether a function $f$ has an inverse and how it's classified can depend on several conditions. We say $f : A to B$ has a left-sided inverse $f^{-1} : B to A$ if $f^{-1} circ f = 1_A$ (which comes from injectivity), a right-sided one if $f circ f^{-1} = 1_B$ (which comes from surjectivity), and a two-sided inverse (or sometimes just "inverse") if $f$ is bijective.
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:50
1
$begingroup$
Now that said, I assume you're just looking to see for left inverses. Context is dependent but I assume you're working with real functions? A valid test is the "horizontal line test" - if you can find a horizontal line that crosses the graph of a function in at least two places, the function is not injective. It doesn't generalize too well, though, and saying "it passes the horizontal line test" is probably not going to work insofar as proofs, but it's at least a good indicator.
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:52
1
$begingroup$
As far as "formally" checking, the notion of $f$ being injective is defined by, if $f(x)=f(x')$, it implies $x=x'$. You can also work with the contrapositive version - if $x neq x'$, then $f(x)neq f(x')$. Which you want to use is partly a matter of personal taste or ease, those are just the ways I personally did so. (And of course, if $f$ is not injective, finding an $x, x'$ with $x neq x'$ such that $f(x)=f(x')$ is 100% valid.)
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:54
add a comment |
$begingroup$
From what I've learnt, a function $f$ has an inverse function $f^{-1}$ if the function $f$ is injective (one-to-one, horizontal rule applies).
How can I check if a function has an inverse in the first place? Given two examples:
1:
$f(x) = frac{arcsin x-1}{arcsin x+2}$
its inverse is:
$f^{-1}(x)= sin frac{-2x-1}{x-1}$
2:
$g(x) = frac{ln x}{x}$
its inverse is:
$g^{-1}(x)= e^{frac{-x}{x-1}}$
But how do I properly check if a function is an injective fuction? Preferably without using derivatives.
Thanks.
real-analysis analysis inverse-function
asked Dec 5 '18 at 2:46
wenoweno
1499
$endgroup$
From what I've learnt, a function $f$ has an inverse function $f^{-1}$ if the function $f$ is injective (one-to-one, horizontal rule applies).
How can I check if a function has an inverse in the first place? Given two examples:
1:
$f(x) = frac{arcsin x-1}{arcsin x+2}$
its inverse is:
$f^{-1}(x)= sin frac{-2x-1}{x-1}$
2:
$g(x) = frac{ln x}{x}$
its inverse is:
$g^{-1}(x)= e^{frac{-x}{x-1}}$
But how do I properly check if a function is an injective fuction? Preferably without using derivatives.
Thanks.
real-analysis analysis inverse-function
real-analysis analysis inverse-function
asked Dec 5 '18 at 2:46
wenoweno
1499
asked Dec 5 '18 at 2:46
wenoweno
1499
asked Dec 5 '18 at 2:46
wenoweno
1499
asked Dec 5 '18 at 2:46
wenoweno
1499
asked Dec 5 '18 at 2:46
wenoweno
1499
1499
1
$begingroup$
Depending on context, I just want to note: whether a function $f$ has an inverse and how it's classified can depend on several conditions. We say $f : A to B$ has a left-sided inverse $f^{-1} : B to A$ if $f^{-1} circ f = 1_A$ (which comes from injectivity), a right-sided one if $f circ f^{-1} = 1_B$ (which comes from surjectivity), and a two-sided inverse (or sometimes just "inverse") if $f$ is bijective.
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:50
1
$begingroup$
Now that said, I assume you're just looking to see for left inverses. Context is dependent but I assume you're working with real functions? A valid test is the "horizontal line test" - if you can find a horizontal line that crosses the graph of a function in at least two places, the function is not injective. It doesn't generalize too well, though, and saying "it passes the horizontal line test" is probably not going to work insofar as proofs, but it's at least a good indicator.
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:52
1
$begingroup$
As far as "formally" checking, the notion of $f$ being injective is defined by, if $f(x)=f(x')$, it implies $x=x'$. You can also work with the contrapositive version - if $x neq x'$, then $f(x)neq f(x')$. Which you want to use is partly a matter of personal taste or ease, those are just the ways I personally did so. (And of course, if $f$ is not injective, finding an $x, x'$ with $x neq x'$ such that $f(x)=f(x')$ is 100% valid.)
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:54
add a comment |
1
$begingroup$
Depending on context, I just want to note: whether a function $f$ has an inverse and how it's classified can depend on several conditions. We say $f : A to B$ has a left-sided inverse $f^{-1} : B to A$ if $f^{-1} circ f = 1_A$ (which comes from injectivity), a right-sided one if $f circ f^{-1} = 1_B$ (which comes from surjectivity), and a two-sided inverse (or sometimes just "inverse") if $f$ is bijective.
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:50
1
$begingroup$
Now that said, I assume you're just looking to see for left inverses. Context is dependent but I assume you're working with real functions? A valid test is the "horizontal line test" - if you can find a horizontal line that crosses the graph of a function in at least two places, the function is not injective. It doesn't generalize too well, though, and saying "it passes the horizontal line test" is probably not going to work insofar as proofs, but it's at least a good indicator.
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:52
1
$begingroup$
As far as "formally" checking, the notion of $f$ being injective is defined by, if $f(x)=f(x')$, it implies $x=x'$. You can also work with the contrapositive version - if $x neq x'$, then $f(x)neq f(x')$. Which you want to use is partly a matter of personal taste or ease, those are just the ways I personally did so. (And of course, if $f$ is not injective, finding an $x, x'$ with $x neq x'$ such that $f(x)=f(x')$ is 100% valid.)
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:54
1
1
$begingroup$
Depending on context, I just want to note: whether a function $f$ has an inverse and how it's classified can depend on several conditions. We say $f : A to B$ has a left-sided inverse $f^{-1} : B to A$ if $f^{-1} circ f = 1_A$ (which comes from injectivity), a right-sided one if $f circ f^{-1} = 1_B$ (which comes from surjectivity), and a two-sided inverse (or sometimes just "inverse") if $f$ is bijective.
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:50
$begingroup$
Depending on context, I just want to note: whether a function $f$ has an inverse and how it's classified can depend on several conditions. We say $f : A to B$ has a left-sided inverse $f^{-1} : B to A$ if $f^{-1} circ f = 1_A$ (which comes from injectivity), a right-sided one if $f circ f^{-1} = 1_B$ (which comes from surjectivity), and a two-sided inverse (or sometimes just "inverse") if $f$ is bijective.
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:50
1
1
$begingroup$
Now that said, I assume you're just looking to see for left inverses. Context is dependent but I assume you're working with real functions? A valid test is the "horizontal line test" - if you can find a horizontal line that crosses the graph of a function in at least two places, the function is not injective. It doesn't generalize too well, though, and saying "it passes the horizontal line test" is probably not going to work insofar as proofs, but it's at least a good indicator.
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:52
$begingroup$
Now that said, I assume you're just looking to see for left inverses. Context is dependent but I assume you're working with real functions? A valid test is the "horizontal line test" - if you can find a horizontal line that crosses the graph of a function in at least two places, the function is not injective. It doesn't generalize too well, though, and saying "it passes the horizontal line test" is probably not going to work insofar as proofs, but it's at least a good indicator.
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:52
1
1
$begingroup$
As far as "formally" checking, the notion of $f$ being injective is defined by, if $f(x)=f(x')$, it implies $x=x'$. You can also work with the contrapositive version - if $x neq x'$, then $f(x)neq f(x')$. Which you want to use is partly a matter of personal taste or ease, those are just the ways I personally did so. (And of course, if $f$ is not injective, finding an $x, x'$ with $x neq x'$ such that $f(x)=f(x')$ is 100% valid.)
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:54
$begingroup$
As far as "formally" checking, the notion of $f$ being injective is defined by, if $f(x)=f(x')$, it implies $x=x'$. You can also work with the contrapositive version - if $x neq x'$, then $f(x)neq f(x')$. Which you want to use is partly a matter of personal taste or ease, those are just the ways I personally did so. (And of course, if $f$ is not injective, finding an $x, x'$ with $x neq x'$ such that $f(x)=f(x')$ is 100% valid.)
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:54
add a comment |
1 Answer
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$begingroup$
If you want to check that a function is injective, of course you need the formal definition
A function $f:Ato B$ is injective if and only if $f(x)=f(y)$ implies $x=y$ for all $x,yin A$.
Let's take your second example, $f:(0,infty)tomathbb R$, where $f(x)=ln x/x$. Supopse $f(x)=f(y)$, that is,
$$ frac{ln x}{x}=frac{ln y}y. $$
This implies
$$yln x = xln y iff xe^y=ye^x.$$
But if you look at the graph of the implicit equation $xe^y=ye^x$, you see that for any $x$ on $(0,infty)$, there is in fact two $y$s in the same range which satisfy the equation. That is, $f(x)=f(y)$ does not imply $x=y$, so this function is not injective.
By the way, it is wrong to say that $f$ has an inverse if and only if it is injective, if we accept ``inverse'' to mean a function $g$ so that $gcirc fequiv fcirc gequiv 1$ where $1$ is the identity mapping. We need instead $f$ to be bijective; i.e. both injective and surjective.
answered Dec 5 '18 at 3:08
YiFanYiFan
2,9411422
$endgroup$
add a comment |
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$begingroup$
If you want to check that a function is injective, of course you need the formal definition
A function $f:Ato B$ is injective if and only if $f(x)=f(y)$ implies $x=y$ for all $x,yin A$.
Let's take your second example, $f:(0,infty)tomathbb R$, where $f(x)=ln x/x$. Supopse $f(x)=f(y)$, that is,
$$ frac{ln x}{x}=frac{ln y}y. $$
This implies
$$yln x = xln y iff xe^y=ye^x.$$
But if you look at the graph of the implicit equation $xe^y=ye^x$, you see that for any $x$ on $(0,infty)$, there is in fact two $y$s in the same range which satisfy the equation. That is, $f(x)=f(y)$ does not imply $x=y$, so this function is not injective.
By the way, it is wrong to say that $f$ has an inverse if and only if it is injective, if we accept ``inverse'' to mean a function $g$ so that $gcirc fequiv fcirc gequiv 1$ where $1$ is the identity mapping. We need instead $f$ to be bijective; i.e. both injective and surjective.
answered Dec 5 '18 at 3:08
YiFanYiFan
2,9411422
$endgroup$
add a comment |
$begingroup$
If you want to check that a function is injective, of course you need the formal definition
A function $f:Ato B$ is injective if and only if $f(x)=f(y)$ implies $x=y$ for all $x,yin A$.
Let's take your second example, $f:(0,infty)tomathbb R$, where $f(x)=ln x/x$. Supopse $f(x)=f(y)$, that is,
$$ frac{ln x}{x}=frac{ln y}y. $$
This implies
$$yln x = xln y iff xe^y=ye^x.$$
But if you look at the graph of the implicit equation $xe^y=ye^x$, you see that for any $x$ on $(0,infty)$, there is in fact two $y$s in the same range which satisfy the equation. That is, $f(x)=f(y)$ does not imply $x=y$, so this function is not injective.
By the way, it is wrong to say that $f$ has an inverse if and only if it is injective, if we accept ``inverse'' to mean a function $g$ so that $gcirc fequiv fcirc gequiv 1$ where $1$ is the identity mapping. We need instead $f$ to be bijective; i.e. both injective and surjective.
answered Dec 5 '18 at 3:08
YiFanYiFan
2,9411422
$endgroup$
add a comment |
$begingroup$
If you want to check that a function is injective, of course you need the formal definition
A function $f:Ato B$ is injective if and only if $f(x)=f(y)$ implies $x=y$ for all $x,yin A$.
Let's take your second example, $f:(0,infty)tomathbb R$, where $f(x)=ln x/x$. Supopse $f(x)=f(y)$, that is,
$$ frac{ln x}{x}=frac{ln y}y. $$
This implies
$$yln x = xln y iff xe^y=ye^x.$$
But if you look at the graph of the implicit equation $xe^y=ye^x$, you see that for any $x$ on $(0,infty)$, there is in fact two $y$s in the same range which satisfy the equation. That is, $f(x)=f(y)$ does not imply $x=y$, so this function is not injective.
By the way, it is wrong to say that $f$ has an inverse if and only if it is injective, if we accept ``inverse'' to mean a function $g$ so that $gcirc fequiv fcirc gequiv 1$ where $1$ is the identity mapping. We need instead $f$ to be bijective; i.e. both injective and surjective.
answered Dec 5 '18 at 3:08
YiFanYiFan
2,9411422
$endgroup$
If you want to check that a function is injective, of course you need the formal definition
A function $f:Ato B$ is injective if and only if $f(x)=f(y)$ implies $x=y$ for all $x,yin A$.
Let's take your second example, $f:(0,infty)tomathbb R$, where $f(x)=ln x/x$. Supopse $f(x)=f(y)$, that is,
$$ frac{ln x}{x}=frac{ln y}y. $$
This implies
$$yln x = xln y iff xe^y=ye^x.$$
But if you look at the graph of the implicit equation $xe^y=ye^x$, you see that for any $x$ on $(0,infty)$, there is in fact two $y$s in the same range which satisfy the equation. That is, $f(x)=f(y)$ does not imply $x=y$, so this function is not injective.
By the way, it is wrong to say that $f$ has an inverse if and only if it is injective, if we accept ``inverse'' to mean a function $g$ so that $gcirc fequiv fcirc gequiv 1$ where $1$ is the identity mapping. We need instead $f$ to be bijective; i.e. both injective and surjective.
answered Dec 5 '18 at 3:08
YiFanYiFan
2,9411422
answered Dec 5 '18 at 3:08
YiFanYiFan
2,9411422
answered Dec 5 '18 at 3:08
YiFanYiFan
2,9411422
answered Dec 5 '18 at 3:08
YiFanYiFan
2,9411422
2,9411422
add a comment |
add a comment |
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1
$begingroup$
Depending on context, I just want to note: whether a function $f$ has an inverse and how it's classified can depend on several conditions. We say $f : A to B$ has a left-sided inverse $f^{-1} : B to A$ if $f^{-1} circ f = 1_A$ (which comes from injectivity), a right-sided one if $f circ f^{-1} = 1_B$ (which comes from surjectivity), and a two-sided inverse (or sometimes just "inverse") if $f$ is bijective.
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:50
1
$begingroup$
Now that said, I assume you're just looking to see for left inverses. Context is dependent but I assume you're working with real functions? A valid test is the "horizontal line test" - if you can find a horizontal line that crosses the graph of a function in at least two places, the function is not injective. It doesn't generalize too well, though, and saying "it passes the horizontal line test" is probably not going to work insofar as proofs, but it's at least a good indicator.
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:52
1
$begingroup$
As far as "formally" checking, the notion of $f$ being injective is defined by, if $f(x)=f(x')$, it implies $x=x'$. You can also work with the contrapositive version - if $x neq x'$, then $f(x)neq f(x')$. Which you want to use is partly a matter of personal taste or ease, those are just the ways I personally did so. (And of course, if $f$ is not injective, finding an $x, x'$ with $x neq x'$ such that $f(x)=f(x')$ is 100% valid.)
$endgroup$
– Eevee Trainer
Dec 5 '18 at 2:54