Just a theory question on odd and even functions

[flyingpig]

Homework Statement

If a function is even or odd, what can one conclude about its inverse?

The Attempt at a Solution

Let f(x) = f(-x)

f^-1(x) = f^-1(-x)

Let g(x) = -g(x)

g^-1(x) = -g^-1(x)

 

[Mark44]

Homework Statement

If a function is even or odd, what can one conclude about its inverse?

The Attempt at a Solution

Let f(x) = f(-x)

f^-1(x) = f^-1(-x)

Let g(x) = -g(x)

g^-1(x) = -g^-1(x)

Let f(x) = x², which is an even function. What can you conclude about its inverse?

 

[flyingpig]

$$\sqrt{x} \neq \sqrt{-x}$$...

I am guessing in general, you can't make any conclusion at all can you lol?

 

[Mark44]

Are you saying that if f(x) = x², f^-1(x) = $\sqrt{x}$?

 

[flyingpig]

Are you saying that if f(x) = x², f^-1(x) = $\sqrt{x}$?

Plus or minus root x

 

[Mark44]

Some questions you should be asking yourself are:
Can an even function have an inverse?
Can an odd function have an inverse?
What sorts of functions have inverses?

 

[Mark44]

Plus or minus root x

Then that's not a function. A function produces a single value for a given input.

 

[flyingpig]

Well okay, odd functions can have an inverse, f(x) = x, f(y) = y

f(-x) = -f(x) = -x

f(-y) = -f(y) = -y

-y = -x

y = x

 

[Mark44]

What about, say, y = x³?

Does it have an inverse? Is x³ its own inverse, like the very simple function you just picked?

If this function has an inverse, is the inverse even, odd, neither?

 

[flyingpig]

It is neither...

So the conclusion is that there are no relationships between odd and even functions and its inverse.

 

[Mark44]

It is neither...

So the conclusion is that there are no relationships between odd and even functions and its inverse.

That's not the right conclusion. You can say this about even functions, because even functions aren't one-to-one, and don't have inverses. An odd function [STRIKE]does[/STRIKE] will have an inverse [STRIKE]because[/STRIKE] if it is one-to-one.

Limiting your focus to odd functions, can you say something about whether their inverses are even or odd?

 

[flyingpig]

No...y = x³. That example you showed me. y = x^1/3 is neither an even nor odd function.

 

[Mark44]

Wrong. y = x^1/3 is an odd function.

 

[LCKurtz]

An odd function does have an inverse, because it is one-to-one.

You mean like sin(x)?

 

[Mark44]

You mean like sin(x)?

I recant what I said. I have edited my original statement.

 

[flyingpig]

One second, I graphed y = x^1/3 on Wolframalpha and it didn't look like it was "anti-symmetric" to me.

I graphed it on Maple and I can only see it on the first quadrant.

 

[Mark44]

Every real number has a cube root, which is the same as saying that the domain of the function y = x^1/3 is all real numbers.

 

[flyingpig]

F(x) = x^1/3

F(-x) = -x^1/3

F(-x) = -F(x)

So its inverse is also odd?

But doesn't F(-x) = -F(x) for x^1/3 also suggest x^1/3 is symmetric about the x-axis and therefore not a function?

 

[Char. Limit]

F(x) = x^1/3

F(-x) = -x^1/3

F(-x) = -F(x)

So its inverse is also odd?

But doesn't F(-x) = -F(x) for x^1/3 also suggest x^1/3 is symmetric about the x-axis and therefore not a function?

Not so. Rather, F(x)=F(-x) implies symmetry about the y-axis, and so if F(x)=F(-x), then F^-1(x) is symmetric about the x-axis. But F(-x)=-F(x) only implies symmetry about the origin.

 

[HallsofIvy]

Yes, that's true- $y= x^{1/3}$ as Mark44 said.

In fact for any odd function, if y= f(x) then f(-x)= -f(x)= -y. That is, if (x, y) is a pair in the function (functions can always be thought of as a set of ordered pairs), then (-x, -y) is also a pair. Now what happens if you reverse the order to (y, x) (getting the set of pairs corresponding to $f^{-1}$)?