What is the maximal domain of these functions?
Need help figuring out, how to find the maximal domains of functions
$f(x) = frac{x^3-x}{2x^2+1}$
$f(x) = frac{x^2-5x+6}{x-2}$
functions
edited Aug 4 '18 at 13:11
pointguard0
1,310821
asked Oct 11 '15 at 17:05
Chris Smith
611
add a comment |
Need help figuring out, how to find the maximal domains of functions
$f(x) = frac{x^3-x}{2x^2+1}$
$f(x) = frac{x^2-5x+6}{x-2}$
functions
edited Aug 4 '18 at 13:11
pointguard0
1,310821
asked Oct 11 '15 at 17:05
Chris Smith
611
I guess $f$ should be dependant on $y$ and $z$, not on $x$? Considering finding the domain: The domain of $f$ is the set of arguments for which $f$ is well-defined. So you need to ask yourself "when is $frac{a}{b}$ not well-defined?".
– Piwi
Oct 11 '15 at 17:28
please clarify the domain of function argument
– pointguard0
Aug 4 '18 at 13:18
add a comment |
Need help figuring out, how to find the maximal domains of functions
$f(x) = frac{x^3-x}{2x^2+1}$
$f(x) = frac{x^2-5x+6}{x-2}$
functions
edited Aug 4 '18 at 13:11
pointguard0
1,310821
asked Oct 11 '15 at 17:05
Chris Smith
611
Need help figuring out, how to find the maximal domains of functions
$f(x) = frac{x^3-x}{2x^2+1}$
$f(x) = frac{x^2-5x+6}{x-2}$
functions
functions
edited Aug 4 '18 at 13:11
pointguard0
1,310821
asked Oct 11 '15 at 17:05
Chris Smith
611
edited Aug 4 '18 at 13:11
pointguard0
1,310821
asked Oct 11 '15 at 17:05
Chris Smith
611
edited Aug 4 '18 at 13:11
pointguard0
1,310821
edited Aug 4 '18 at 13:11
pointguard0
1,310821
edited Aug 4 '18 at 13:11
pointguard0
1,310821
1,310821
asked Oct 11 '15 at 17:05
Chris Smith
611
asked Oct 11 '15 at 17:05
Chris Smith
611
asked Oct 11 '15 at 17:05
Chris Smith
611
611
I guess $f$ should be dependant on $y$ and $z$, not on $x$? Considering finding the domain: The domain of $f$ is the set of arguments for which $f$ is well-defined. So you need to ask yourself "when is $frac{a}{b}$ not well-defined?".
– Piwi
Oct 11 '15 at 17:28
please clarify the domain of function argument
– pointguard0
Aug 4 '18 at 13:18
add a comment |
I guess $f$ should be dependant on $y$ and $z$, not on $x$? Considering finding the domain: The domain of $f$ is the set of arguments for which $f$ is well-defined. So you need to ask yourself "when is $frac{a}{b}$ not well-defined?".
– Piwi
Oct 11 '15 at 17:28
please clarify the domain of function argument
– pointguard0
Aug 4 '18 at 13:18
I guess $f$ should be dependant on $y$ and $z$, not on $x$? Considering finding the domain: The domain of $f$ is the set of arguments for which $f$ is well-defined. So you need to ask yourself "when is $frac{a}{b}$ not well-defined?".
– Piwi
Oct 11 '15 at 17:28
I guess $f$ should be dependant on $y$ and $z$, not on $x$? Considering finding the domain: The domain of $f$ is the set of arguments for which $f$ is well-defined. So you need to ask yourself "when is $frac{a}{b}$ not well-defined?".
– Piwi
Oct 11 '15 at 17:28
please clarify the domain of function argument
– pointguard0
Aug 4 '18 at 13:18
please clarify the domain of function argument
– pointguard0
Aug 4 '18 at 13:18
add a comment |
2 Answers
2
active
oldest
votes
I will suppose $f(x)$ should be $f(y)$ and $f(z)$ and that we are working in $mathbb{R}$.
This is a rational expression. The only restriction is that the denominator is not $0$. Since
$$2y^2+1=0$$
has no solution then the denominator does not impose any restriction on the domain. Hence $dom(f)=mathbb{R}$.Same thinking leads to
$$z-2ne 0 Leftrightarrow zne 2.$$
The domain is $mathbb{R}backslash{2}$.
answered Oct 11 '15 at 22:22
Jean-François Gagnon
897315
function argument could be complex.
– pointguard0
Aug 4 '18 at 13:04
add a comment |
The second function does not contain $z=2$ in the domain, but it can be extended continuously at this point, thus defined continuous, so the maximal domain would be all real numbers.
answered Oct 11 '15 at 23:47
user279325
17416
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
I will suppose $f(x)$ should be $f(y)$ and $f(z)$ and that we are working in $mathbb{R}$.
This is a rational expression. The only restriction is that the denominator is not $0$. Since
$$2y^2+1=0$$
has no solution then the denominator does not impose any restriction on the domain. Hence $dom(f)=mathbb{R}$.Same thinking leads to
$$z-2ne 0 Leftrightarrow zne 2.$$
The domain is $mathbb{R}backslash{2}$.
answered Oct 11 '15 at 22:22
Jean-François Gagnon
897315
function argument could be complex.
– pointguard0
Aug 4 '18 at 13:04
add a comment |
I will suppose $f(x)$ should be $f(y)$ and $f(z)$ and that we are working in $mathbb{R}$.
This is a rational expression. The only restriction is that the denominator is not $0$. Since
$$2y^2+1=0$$
has no solution then the denominator does not impose any restriction on the domain. Hence $dom(f)=mathbb{R}$.Same thinking leads to
$$z-2ne 0 Leftrightarrow zne 2.$$
The domain is $mathbb{R}backslash{2}$.
answered Oct 11 '15 at 22:22
Jean-François Gagnon
897315
function argument could be complex.
– pointguard0
Aug 4 '18 at 13:04
add a comment |
I will suppose $f(x)$ should be $f(y)$ and $f(z)$ and that we are working in $mathbb{R}$.
This is a rational expression. The only restriction is that the denominator is not $0$. Since
$$2y^2+1=0$$
has no solution then the denominator does not impose any restriction on the domain. Hence $dom(f)=mathbb{R}$.Same thinking leads to
$$z-2ne 0 Leftrightarrow zne 2.$$
The domain is $mathbb{R}backslash{2}$.
answered Oct 11 '15 at 22:22
Jean-François Gagnon
897315
I will suppose $f(x)$ should be $f(y)$ and $f(z)$ and that we are working in $mathbb{R}$.
This is a rational expression. The only restriction is that the denominator is not $0$. Since
$$2y^2+1=0$$
has no solution then the denominator does not impose any restriction on the domain. Hence $dom(f)=mathbb{R}$.Same thinking leads to
$$z-2ne 0 Leftrightarrow zne 2.$$
The domain is $mathbb{R}backslash{2}$.
answered Oct 11 '15 at 22:22
Jean-François Gagnon
897315
answered Oct 11 '15 at 22:22
Jean-François Gagnon
897315
answered Oct 11 '15 at 22:22
Jean-François Gagnon
897315
answered Oct 11 '15 at 22:22
Jean-François Gagnon
897315
897315
function argument could be complex.
– pointguard0
Aug 4 '18 at 13:04
add a comment |
function argument could be complex.
– pointguard0
Aug 4 '18 at 13:04
function argument could be complex.
– pointguard0
Aug 4 '18 at 13:04
function argument could be complex.
– pointguard0
Aug 4 '18 at 13:04
add a comment |
The second function does not contain $z=2$ in the domain, but it can be extended continuously at this point, thus defined continuous, so the maximal domain would be all real numbers.
answered Oct 11 '15 at 23:47
user279325
17416
add a comment |
The second function does not contain $z=2$ in the domain, but it can be extended continuously at this point, thus defined continuous, so the maximal domain would be all real numbers.
answered Oct 11 '15 at 23:47
user279325
17416
add a comment |
The second function does not contain $z=2$ in the domain, but it can be extended continuously at this point, thus defined continuous, so the maximal domain would be all real numbers.
answered Oct 11 '15 at 23:47
user279325
17416
The second function does not contain $z=2$ in the domain, but it can be extended continuously at this point, thus defined continuous, so the maximal domain would be all real numbers.
answered Oct 11 '15 at 23:47
user279325
17416
answered Oct 11 '15 at 23:47
user279325
17416
answered Oct 11 '15 at 23:47
user279325
17416
answered Oct 11 '15 at 23:47
user279325
17416
17416
add a comment |
add a comment |
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I guess $f$ should be dependant on $y$ and $z$, not on $x$? Considering finding the domain: The domain of $f$ is the set of arguments for which $f$ is well-defined. So you need to ask yourself "when is $frac{a}{b}$ not well-defined?".
– Piwi
Oct 11 '15 at 17:28
please clarify the domain of function argument
– pointguard0
Aug 4 '18 at 13:18