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lLovePhysics
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How is y=0 an odd function when it isn't symmetric to the origin?
Did you mean "odd" and "even"? f(x)= 0 is clearly a closed function, certainly not open!CompuChip said:Note here, that for any function f in general, it's not necessary that f is either closed or open; nor that it cannot be both. You've found an example of the latter
An odd function is a type of mathematical function where the output of the function changes sign when the input is changed to its negative value. In other words, if f(x) is an odd function, then f(-x) = -f(x).
When y=0, the output of the function is always zero regardless of the input. This means that for any value of x, f(-x) = f(x) = 0. Since the sign of 0 remains the same when the input is changed to its negative value, y=0 can be considered an odd function.
The significance of y=0 being an odd function is that it helps us identify symmetry in graphs. An odd function has a specific type of symmetry called origin symmetry, which means that the graph is symmetric about the origin (0,0). This can be seen by the fact that when the function is reflected about the origin, the graph remains unchanged.
To prove that y=0 is an odd function, we can use the definition of an odd function and show that f(-x) = -f(x). Since y=0 is a constant function, f(-x) = 0 and -f(x) = 0, which means that f(-x) = -f(x) is true for all values of x. Therefore, y=0 is an odd function.
No, a function cannot be both even and odd. A function is considered even if f(x) = f(-x) for all values of x, while a function is considered odd if f(-x) = -f(x) for all values of x. These two conditions cannot be satisfied simultaneously, so a function cannot be both even and odd.