Function Transformations - how do I find the invariant points?
1
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When they give you a picture of a graph (doesn't matter what kind - linear, parabola, inverse, etc.) how do you find the invariant points?
functions
asked Jun 13 '15 at 17:08
KatKat
6113
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add a comment |
1
$begingroup$
When they give you a picture of a graph (doesn't matter what kind - linear, parabola, inverse, etc.) how do you find the invariant points?
functions
asked Jun 13 '15 at 17:08
KatKat
6113
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$begingroup$
Possibly related: Analysis Question? Fixed Point? and Fixed point location for functions
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– epimorphic
Jun 13 '15 at 17:35
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I believe the question is how to determine invariant points between a function and it's inverse. If you're looking to algebraically find the point, you just make the two functions equal each other, and then solve for x.
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– Rock
Dec 15 '17 at 2:33
add a comment |
1
1
1
$begingroup$
When they give you a picture of a graph (doesn't matter what kind - linear, parabola, inverse, etc.) how do you find the invariant points?
functions
asked Jun 13 '15 at 17:08
KatKat
6113
$endgroup$
When they give you a picture of a graph (doesn't matter what kind - linear, parabola, inverse, etc.) how do you find the invariant points?
functions
functions
asked Jun 13 '15 at 17:08
KatKat
6113
asked Jun 13 '15 at 17:08
KatKat
6113
asked Jun 13 '15 at 17:08
KatKat
6113
asked Jun 13 '15 at 17:08
KatKat
6113
asked Jun 13 '15 at 17:08
KatKat
6113
6113
$begingroup$
Possibly related: Analysis Question? Fixed Point? and Fixed point location for functions
$endgroup$
– epimorphic
Jun 13 '15 at 17:35
$begingroup$
I believe the question is how to determine invariant points between a function and it's inverse. If you're looking to algebraically find the point, you just make the two functions equal each other, and then solve for x.
$endgroup$
– Rock
Dec 15 '17 at 2:33
add a comment |
$begingroup$
Possibly related: Analysis Question? Fixed Point? and Fixed point location for functions
$endgroup$
– epimorphic
Jun 13 '15 at 17:35
$begingroup$
I believe the question is how to determine invariant points between a function and it's inverse. If you're looking to algebraically find the point, you just make the two functions equal each other, and then solve for x.
$endgroup$
– Rock
Dec 15 '17 at 2:33
$begingroup$
Possibly related: Analysis Question? Fixed Point? and Fixed point location for functions
$endgroup$
– epimorphic
Jun 13 '15 at 17:35
$begingroup$
Possibly related: Analysis Question? Fixed Point? and Fixed point location for functions
$endgroup$
– epimorphic
Jun 13 '15 at 17:35
$begingroup$
I believe the question is how to determine invariant points between a function and it's inverse. If you're looking to algebraically find the point, you just make the two functions equal each other, and then solve for x.
$endgroup$
– Rock
Dec 15 '17 at 2:33
$begingroup$
I believe the question is how to determine invariant points between a function and it's inverse. If you're looking to algebraically find the point, you just make the two functions equal each other, and then solve for x.
$endgroup$
– Rock
Dec 15 '17 at 2:33
add a comment |
1 Answer
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2
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You're looking for points where $f(x)=x$. Phrased differently, you're looking for points where $(x,x)$ lies on your graph. Rephrased once more, you're looking for intersection with the line $y=x$.
answered Jun 13 '15 at 17:12
Zach StoneZach Stone
4,62231623
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$begingroup$
I get what I am looking for but how do I find it?
$endgroup$
– Kat
Jun 13 '15 at 17:17
1
$begingroup$
Draw the line $y=x$, and look for intersection with the given graph.
$endgroup$
– Zach Stone
Jun 13 '15 at 17:24
$begingroup$
Is there any way to do it algebraically?
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– Kat
Jun 13 '15 at 17:28
1
$begingroup$
Wait, are you given a picture or equation? If you have the equation, set $f(x)=x$ and try to solve for $x$. I can't be more specific without more details about what $f$ looks like. Sometimes it's possible, sometimes it's not.
$endgroup$
– Zach Stone
Jun 13 '15 at 17:30
$begingroup$
It is just a picture of a graph of two porabolas side by side (one is a reflection over the y-intercept) that will eventually intersect but do not in the picture.
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– Kat
Jun 13 '15 at 17:42
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show 1 more comment
protected by Community♦ Jan 10 at 6:56
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
You're looking for points where $f(x)=x$. Phrased differently, you're looking for points where $(x,x)$ lies on your graph. Rephrased once more, you're looking for intersection with the line $y=x$.
answered Jun 13 '15 at 17:12
Zach StoneZach Stone
4,62231623
$endgroup$
$begingroup$
I get what I am looking for but how do I find it?
$endgroup$
– Kat
Jun 13 '15 at 17:17
1
$begingroup$
Draw the line $y=x$, and look for intersection with the given graph.
$endgroup$
– Zach Stone
Jun 13 '15 at 17:24
$begingroup$
Is there any way to do it algebraically?
$endgroup$
– Kat
Jun 13 '15 at 17:28
1
$begingroup$
Wait, are you given a picture or equation? If you have the equation, set $f(x)=x$ and try to solve for $x$. I can't be more specific without more details about what $f$ looks like. Sometimes it's possible, sometimes it's not.
$endgroup$
– Zach Stone
Jun 13 '15 at 17:30
$begingroup$
It is just a picture of a graph of two porabolas side by side (one is a reflection over the y-intercept) that will eventually intersect but do not in the picture.
$endgroup$
– Kat
Jun 13 '15 at 17:42
|
show 1 more comment
2
$begingroup$
You're looking for points where $f(x)=x$. Phrased differently, you're looking for points where $(x,x)$ lies on your graph. Rephrased once more, you're looking for intersection with the line $y=x$.
answered Jun 13 '15 at 17:12
Zach StoneZach Stone
4,62231623
$endgroup$
$begingroup$
I get what I am looking for but how do I find it?
$endgroup$
– Kat
Jun 13 '15 at 17:17
1
$begingroup$
Draw the line $y=x$, and look for intersection with the given graph.
$endgroup$
– Zach Stone
Jun 13 '15 at 17:24
$begingroup$
Is there any way to do it algebraically?
$endgroup$
– Kat
Jun 13 '15 at 17:28
1
$begingroup$
Wait, are you given a picture or equation? If you have the equation, set $f(x)=x$ and try to solve for $x$. I can't be more specific without more details about what $f$ looks like. Sometimes it's possible, sometimes it's not.
$endgroup$
– Zach Stone
Jun 13 '15 at 17:30
$begingroup$
It is just a picture of a graph of two porabolas side by side (one is a reflection over the y-intercept) that will eventually intersect but do not in the picture.
$endgroup$
– Kat
Jun 13 '15 at 17:42
|
show 1 more comment
2
2
2
$begingroup$
You're looking for points where $f(x)=x$. Phrased differently, you're looking for points where $(x,x)$ lies on your graph. Rephrased once more, you're looking for intersection with the line $y=x$.
answered Jun 13 '15 at 17:12
Zach StoneZach Stone
4,62231623
$endgroup$
You're looking for points where $f(x)=x$. Phrased differently, you're looking for points where $(x,x)$ lies on your graph. Rephrased once more, you're looking for intersection with the line $y=x$.
answered Jun 13 '15 at 17:12
Zach StoneZach Stone
4,62231623
answered Jun 13 '15 at 17:12
Zach StoneZach Stone
4,62231623
answered Jun 13 '15 at 17:12
Zach StoneZach Stone
4,62231623
answered Jun 13 '15 at 17:12
Zach StoneZach Stone
4,62231623
4,62231623
$begingroup$
I get what I am looking for but how do I find it?
$endgroup$
– Kat
Jun 13 '15 at 17:17
1
$begingroup$
Draw the line $y=x$, and look for intersection with the given graph.
$endgroup$
– Zach Stone
Jun 13 '15 at 17:24
$begingroup$
Is there any way to do it algebraically?
$endgroup$
– Kat
Jun 13 '15 at 17:28
1
$begingroup$
Wait, are you given a picture or equation? If you have the equation, set $f(x)=x$ and try to solve for $x$. I can't be more specific without more details about what $f$ looks like. Sometimes it's possible, sometimes it's not.
$endgroup$
– Zach Stone
Jun 13 '15 at 17:30
$begingroup$
It is just a picture of a graph of two porabolas side by side (one is a reflection over the y-intercept) that will eventually intersect but do not in the picture.
$endgroup$
– Kat
Jun 13 '15 at 17:42
|
show 1 more comment
$begingroup$
I get what I am looking for but how do I find it?
$endgroup$
– Kat
Jun 13 '15 at 17:17
1
$begingroup$
Draw the line $y=x$, and look for intersection with the given graph.
$endgroup$
– Zach Stone
Jun 13 '15 at 17:24
$begingroup$
Is there any way to do it algebraically?
$endgroup$
– Kat
Jun 13 '15 at 17:28
1
$begingroup$
Wait, are you given a picture or equation? If you have the equation, set $f(x)=x$ and try to solve for $x$. I can't be more specific without more details about what $f$ looks like. Sometimes it's possible, sometimes it's not.
$endgroup$
– Zach Stone
Jun 13 '15 at 17:30
$begingroup$
It is just a picture of a graph of two porabolas side by side (one is a reflection over the y-intercept) that will eventually intersect but do not in the picture.
$endgroup$
– Kat
Jun 13 '15 at 17:42
$begingroup$
I get what I am looking for but how do I find it?
$endgroup$
– Kat
Jun 13 '15 at 17:17
$begingroup$
I get what I am looking for but how do I find it?
$endgroup$
– Kat
Jun 13 '15 at 17:17
1
1
$begingroup$
Draw the line $y=x$, and look for intersection with the given graph.
$endgroup$
– Zach Stone
Jun 13 '15 at 17:24
$begingroup$
Draw the line $y=x$, and look for intersection with the given graph.
$endgroup$
– Zach Stone
Jun 13 '15 at 17:24
$begingroup$
Is there any way to do it algebraically?
$endgroup$
– Kat
Jun 13 '15 at 17:28
$begingroup$
Is there any way to do it algebraically?
$endgroup$
– Kat
Jun 13 '15 at 17:28
1
1
$begingroup$
Wait, are you given a picture or equation? If you have the equation, set $f(x)=x$ and try to solve for $x$. I can't be more specific without more details about what $f$ looks like. Sometimes it's possible, sometimes it's not.
$endgroup$
– Zach Stone
Jun 13 '15 at 17:30
$begingroup$
Wait, are you given a picture or equation? If you have the equation, set $f(x)=x$ and try to solve for $x$. I can't be more specific without more details about what $f$ looks like. Sometimes it's possible, sometimes it's not.
$endgroup$
– Zach Stone
Jun 13 '15 at 17:30
$begingroup$
It is just a picture of a graph of two porabolas side by side (one is a reflection over the y-intercept) that will eventually intersect but do not in the picture.
$endgroup$
– Kat
Jun 13 '15 at 17:42
$begingroup$
It is just a picture of a graph of two porabolas side by side (one is a reflection over the y-intercept) that will eventually intersect but do not in the picture.
$endgroup$
– Kat
Jun 13 '15 at 17:42
|
show 1 more comment
protected by Community♦ Jan 10 at 6:56
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
$begingroup$
Possibly related: Analysis Question? Fixed Point? and Fixed point location for functions
$endgroup$
– epimorphic
Jun 13 '15 at 17:35
$begingroup$
I believe the question is how to determine invariant points between a function and it's inverse. If you're looking to algebraically find the point, you just make the two functions equal each other, and then solve for x.
$endgroup$
– Rock
Dec 15 '17 at 2:33