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


by hb20007
Tags: axis, functions, reflection, transformations
hb20007
hb20007 is offline
#1
Dec6-13, 01:20 PM
hb20007's Avatar
P: 18
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?
Phys.Org News Partner Mathematics news on Phys.org
Math modeling handbook now available
Hyperbolic homogeneous polynomials, oh my!
Researchers help Boston Marathon organizers plan for 2014 race
Shyan
Shyan is online now
#2
Dec6-13, 01:38 PM
Shyan's Avatar
P: 740
Its [itex] f^{-1}(x) [/itex]
Very beautiful!
Mark44
Mark44 is online now
#3
Dec6-13, 02:17 PM
Mentor
P: 21,069
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
Shyan is online now
#4
Dec6-13, 02:41 PM
Shyan's Avatar
P: 740

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
Mark44 is online now
#5
Dec6-13, 03:20 PM
Mentor
P: 21,069
Understood. My point was that you can't refer to it as f-1(x).
R136a1
R136a1 is offline
#6
Dec6-13, 07:48 PM
Newcomer
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
hb20007 is offline
#7
Dec7-13, 04:19 AM
hb20007's Avatar
P: 18
Okay, now how about a reflection over y = -x?
Shyan
Shyan is online now
#8
Dec7-13, 08:19 AM
Shyan's Avatar
P: 740
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
hb20007 is offline
#9
Dec7-13, 11:22 AM
hb20007's Avatar
P: 18
Yeah, makes sense...
Thanks


Register to reply

Related Discussions
What is the difference between canonical transformations and gauge transformations? Classical Physics 1
Conformal transformations and Möbius transformations Differential Geometry 7
Classical canonical transformations and unitary transformations in quantum mechanics Quantum Physics 21
transformations Calculus & Beyond Homework 3
Gauge Transformations and (Generalized) Bogoliubov Transformations. Quantum Physics 8