Why does simplifying a function change its domain?
$begingroup$
Perhaps this is a silly question, but if you have a function, such as
$$f(x) = frac{x^2}{x}$$
the domain is all real numbers except x = 0.
However, this function simplifies to
$$f(x) = x$$
which has a domain of all real numbers. The domains for the two functions are different. Why are you permitted to simplify the first function if the domain changes?
functions algebras
asked May 25 '14 at 3:17
JackJack
19110
$endgroup$
add a comment |
$begingroup$
Perhaps this is a silly question, but if you have a function, such as
$$f(x) = frac{x^2}{x}$$
the domain is all real numbers except x = 0.
However, this function simplifies to
$$f(x) = x$$
which has a domain of all real numbers. The domains for the two functions are different. Why are you permitted to simplify the first function if the domain changes?
functions algebras
asked May 25 '14 at 3:17
JackJack
19110
$endgroup$
$begingroup$
To be short, the (maximum valid) domain changes because the reason why the domain is restricted falls away.
$endgroup$
– glglgl
May 25 '14 at 5:54
2
$begingroup$
Possible duplicate of Why does factoring eliminate a hole in the limit?
$endgroup$
– Alex M.
Sep 14 '16 at 21:56
add a comment |
$begingroup$
Perhaps this is a silly question, but if you have a function, such as
$$f(x) = frac{x^2}{x}$$
the domain is all real numbers except x = 0.
However, this function simplifies to
$$f(x) = x$$
which has a domain of all real numbers. The domains for the two functions are different. Why are you permitted to simplify the first function if the domain changes?
functions algebras
asked May 25 '14 at 3:17
JackJack
19110
$endgroup$
Perhaps this is a silly question, but if you have a function, such as
$$f(x) = frac{x^2}{x}$$
the domain is all real numbers except x = 0.
However, this function simplifies to
$$f(x) = x$$
which has a domain of all real numbers. The domains for the two functions are different. Why are you permitted to simplify the first function if the domain changes?
functions algebras
functions algebras
asked May 25 '14 at 3:17
JackJack
19110
asked May 25 '14 at 3:17
JackJack
19110
asked May 25 '14 at 3:17
JackJack
19110
asked May 25 '14 at 3:17
JackJack
19110
asked May 25 '14 at 3:17
JackJack
19110
19110
$begingroup$
To be short, the (maximum valid) domain changes because the reason why the domain is restricted falls away.
$endgroup$
– glglgl
May 25 '14 at 5:54
2
$begingroup$
Possible duplicate of Why does factoring eliminate a hole in the limit?
$endgroup$
– Alex M.
Sep 14 '16 at 21:56
add a comment |
$begingroup$
To be short, the (maximum valid) domain changes because the reason why the domain is restricted falls away.
$endgroup$
– glglgl
May 25 '14 at 5:54
2
$begingroup$
Possible duplicate of Why does factoring eliminate a hole in the limit?
$endgroup$
– Alex M.
Sep 14 '16 at 21:56
$begingroup$
To be short, the (maximum valid) domain changes because the reason why the domain is restricted falls away.
$endgroup$
– glglgl
May 25 '14 at 5:54
$begingroup$
To be short, the (maximum valid) domain changes because the reason why the domain is restricted falls away.
$endgroup$
– glglgl
May 25 '14 at 5:54
2
2
$begingroup$
Possible duplicate of Why does factoring eliminate a hole in the limit?
$endgroup$
– Alex M.
Sep 14 '16 at 21:56
$begingroup$
Possible duplicate of Why does factoring eliminate a hole in the limit?
$endgroup$
– Alex M.
Sep 14 '16 at 21:56
add a comment |
2 Answers
2
active
oldest
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$begingroup$
To be strictly correct, the domain does not change. The simplified version of
$$f(x)=frac{x^2}{x} ,quad xne0$$
is
$$f(x)=x ,quad xne0 .$$
Really, a function is not properly specified unless the domain is stated. If a function is given without a domain then the normal default assumption is to take the largest possible domain. So, if you were given a proposed function $f(x)=x$ with no "back story" you would probably (but need not) choose the domain $Bbb R$. But this case is different since you do have some prior information about the function.
answered May 25 '14 at 3:33
DavidDavid
69.4k667130
$endgroup$
add a comment |
$begingroup$
Think of domain of more as a graph. If you have $y=x$ (as in your example), is the function defined at $0$? yes. If you have $y=frac{x^2}{x}$ and you plug in $0$, it's an improper fraction, so it wouldn't even "work" unless it was simplified.
If you have something like $frac{(x-3)}{(x^2-9)}$, $x$ can't be $3$, right? But this is a "legit" function at $x=3$ (undefined, rather than improper). So when you simplify something like this to $y=frac{1}{(x+3)}$ the domain is STILL $x neq 3$
edited May 26 '14 at 22:48
user7000
2791822
answered May 25 '14 at 5:22
user121955user121955
1487
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
To be strictly correct, the domain does not change. The simplified version of
$$f(x)=frac{x^2}{x} ,quad xne0$$
is
$$f(x)=x ,quad xne0 .$$
Really, a function is not properly specified unless the domain is stated. If a function is given without a domain then the normal default assumption is to take the largest possible domain. So, if you were given a proposed function $f(x)=x$ with no "back story" you would probably (but need not) choose the domain $Bbb R$. But this case is different since you do have some prior information about the function.
answered May 25 '14 at 3:33
DavidDavid
69.4k667130
$endgroup$
add a comment |
$begingroup$
To be strictly correct, the domain does not change. The simplified version of
$$f(x)=frac{x^2}{x} ,quad xne0$$
is
$$f(x)=x ,quad xne0 .$$
Really, a function is not properly specified unless the domain is stated. If a function is given without a domain then the normal default assumption is to take the largest possible domain. So, if you were given a proposed function $f(x)=x$ with no "back story" you would probably (but need not) choose the domain $Bbb R$. But this case is different since you do have some prior information about the function.
answered May 25 '14 at 3:33
DavidDavid
69.4k667130
$endgroup$
add a comment |
$begingroup$
To be strictly correct, the domain does not change. The simplified version of
$$f(x)=frac{x^2}{x} ,quad xne0$$
is
$$f(x)=x ,quad xne0 .$$
Really, a function is not properly specified unless the domain is stated. If a function is given without a domain then the normal default assumption is to take the largest possible domain. So, if you were given a proposed function $f(x)=x$ with no "back story" you would probably (but need not) choose the domain $Bbb R$. But this case is different since you do have some prior information about the function.
answered May 25 '14 at 3:33
DavidDavid
69.4k667130
$endgroup$
To be strictly correct, the domain does not change. The simplified version of
$$f(x)=frac{x^2}{x} ,quad xne0$$
is
$$f(x)=x ,quad xne0 .$$
Really, a function is not properly specified unless the domain is stated. If a function is given without a domain then the normal default assumption is to take the largest possible domain. So, if you were given a proposed function $f(x)=x$ with no "back story" you would probably (but need not) choose the domain $Bbb R$. But this case is different since you do have some prior information about the function.
answered May 25 '14 at 3:33
DavidDavid
69.4k667130
answered May 25 '14 at 3:33
DavidDavid
69.4k667130
answered May 25 '14 at 3:33
DavidDavid
69.4k667130
answered May 25 '14 at 3:33
DavidDavid
69.4k667130
69.4k667130
add a comment |
add a comment |
$begingroup$
Think of domain of more as a graph. If you have $y=x$ (as in your example), is the function defined at $0$? yes. If you have $y=frac{x^2}{x}$ and you plug in $0$, it's an improper fraction, so it wouldn't even "work" unless it was simplified.
If you have something like $frac{(x-3)}{(x^2-9)}$, $x$ can't be $3$, right? But this is a "legit" function at $x=3$ (undefined, rather than improper). So when you simplify something like this to $y=frac{1}{(x+3)}$ the domain is STILL $x neq 3$
edited May 26 '14 at 22:48
user7000
2791822
answered May 25 '14 at 5:22
user121955user121955
1487
$endgroup$
add a comment |
$begingroup$
Think of domain of more as a graph. If you have $y=x$ (as in your example), is the function defined at $0$? yes. If you have $y=frac{x^2}{x}$ and you plug in $0$, it's an improper fraction, so it wouldn't even "work" unless it was simplified.
If you have something like $frac{(x-3)}{(x^2-9)}$, $x$ can't be $3$, right? But this is a "legit" function at $x=3$ (undefined, rather than improper). So when you simplify something like this to $y=frac{1}{(x+3)}$ the domain is STILL $x neq 3$
edited May 26 '14 at 22:48
user7000
2791822
answered May 25 '14 at 5:22
user121955user121955
1487
$endgroup$
add a comment |
$begingroup$
Think of domain of more as a graph. If you have $y=x$ (as in your example), is the function defined at $0$? yes. If you have $y=frac{x^2}{x}$ and you plug in $0$, it's an improper fraction, so it wouldn't even "work" unless it was simplified.
If you have something like $frac{(x-3)}{(x^2-9)}$, $x$ can't be $3$, right? But this is a "legit" function at $x=3$ (undefined, rather than improper). So when you simplify something like this to $y=frac{1}{(x+3)}$ the domain is STILL $x neq 3$
edited May 26 '14 at 22:48
user7000
2791822
answered May 25 '14 at 5:22
user121955user121955
1487
$endgroup$
Think of domain of more as a graph. If you have $y=x$ (as in your example), is the function defined at $0$? yes. If you have $y=frac{x^2}{x}$ and you plug in $0$, it's an improper fraction, so it wouldn't even "work" unless it was simplified.
If you have something like $frac{(x-3)}{(x^2-9)}$, $x$ can't be $3$, right? But this is a "legit" function at $x=3$ (undefined, rather than improper). So when you simplify something like this to $y=frac{1}{(x+3)}$ the domain is STILL $x neq 3$
edited May 26 '14 at 22:48
user7000
2791822
answered May 25 '14 at 5:22
user121955user121955
1487
edited May 26 '14 at 22:48
user7000
2791822
edited May 26 '14 at 22:48
user7000
2791822
edited May 26 '14 at 22:48
user7000
2791822
2791822
answered May 25 '14 at 5:22
user121955user121955
1487
answered May 25 '14 at 5:22
user121955user121955
1487
answered May 25 '14 at 5:22
user121955user121955
1487
1487
add a comment |
add a comment |
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$begingroup$
To be short, the (maximum valid) domain changes because the reason why the domain is restricted falls away.
$endgroup$
– glglgl
May 25 '14 at 5:54
2
$begingroup$
Possible duplicate of Why does factoring eliminate a hole in the limit?
$endgroup$
– Alex M.
Sep 14 '16 at 21:56