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F(-x) is a reflection over the y axis -f(x)

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hb20007
#1
Dec6-13, 01:20 PM
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f(-x) is a reflection over the y axis
-f(x) is a reflection over the x axis

Now, how do we represent a reflection over y=x?
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Shyan
#2
Dec6-13, 01:38 PM
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Its [itex] f^{-1}(x) [/itex]
Very beautiful!
Mark44
#3
Dec6-13, 02:17 PM
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Quote Quote by hb20007 View Post
f(-x) is a reflection over the y axis
-f(x) is a reflection over the x axis

Now, how do we represent a reflection over y=x?
If (x, y) is a point on the graph of f, (y, x) will be the reflection of that point across the line y = x.

Quote Quote by Shyan View Post
Its [itex] f^{-1}(x) [/itex]
Very beautiful!
What if f doesn't have an inverse? For example, y = f(x) = x2. This function is not one-to-one, so doesn't have an inverse.

Shyan
#4
Dec6-13, 02:41 PM
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F(-x) is a reflection over the y axis -f(x)

Quote Quote by Mark44 View Post
If (x, y) is a point on the graph of f, (y, x) will be the reflection of that point across the line y = x.


What if f doesn't have an inverse? For example, y = f(x) = x2. This function is not one-to-one, so doesn't have an inverse.
If a function is not one to one,then there is no function that is its inverse.But there is of course a relation which is the function's inverse.And that relation can be ploted.For [itex]y=x^2 [/itex] we have [itex] x=\pm \sqrt{y} [/itex]which is a two-valued relation between x and y.
Mark44
#5
Dec6-13, 03:20 PM
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Understood. My point was that you can't refer to it as f-1(x).
R136a1
#6
Dec6-13, 07:48 PM
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P: 341
Quote Quote by Mark44 View Post
If (x, y) is a point on the graph of f, (y, x) will be the reflection of that point across the line y = x.


What if f doesn't have an inverse? For example, y = f(x) = x2. This function is not one-to-one, so doesn't have an inverse.
Every function is a relation. If ##R## is a relation, then ##R^{-1}## is a well-defined relation.
hb20007
#7
Dec7-13, 04:19 AM
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Okay, now how about a reflection over y = -x?
Shyan
#8
Dec7-13, 08:19 AM
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Let's see...a reflection over line y=-x means [itex] (x_0,y_0)\rightarrow(-y_0,-x_0) [/itex].
It think it should be [itex]-f^{-1}(-x)[/itex]...ohh...sorry...[itex]-R^{-1}(-x) [/itex].
hb20007
#9
Dec7-13, 11:22 AM
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P: 18
Yeah, makes sense...
Thanks


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