Questions

Will the inverse of an odd function always be a function?

Will the inverse of an odd function always be a function?

The inverse of an odd function is odd (e.g. arctan(x) is odd as tan(x) is odd). f(x) + f(−x) for any function f(x). Hence ex + e-x is even.

Is the inverse of an even function also an even function?

Even functions have graphs that are symmetric with respect to the y-axis. So, if (x,y) is on the graph, then (-x, y) is also on the graph. Consequently, even functions are not one-to -one, and therefore do not have inverses.

How do you prove that the inverse of a function is even or odd?

If f(x) is an odd function and if f(x)=k|x∈I, then:

  1. f(−x)=−f(x)=−k.
  2. −f−1(k)=−x=f−1(−k)
  3. f−1(−k)=−f−1(k)
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Can a function be odd if it doesn’t go through the origin?

If an odd function is defined at zero, then its graph must pass through the origin.

Can a function be both even and odd at the same time?

Can an equation be both even and odd? The only function which is both even and odd is f(x) = 0, defined for all real numbers. This is just a line which sits on the x-axis. If you count equations which are not a function in terms of y, then x=0 would also be both even and odd, and is just a line on the y-axis.

Is a function odd or even?

If you end up with the exact same function that you started with (that is, if f (–x) = f (x), so all of the signs are the same), then the function is even. If you end up with the exact opposite of what you started with (that is, if f (–x) = –f (x), so all of the signs are switched), then the function is odd.

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Is function odd or even?

You may be asked to “determine algebraically” whether a function is even or odd. To do this, you take the function and plug –x in for x, and then simplify. If you end up with the exact opposite of what you started with (that is, if f (–x) = –f (x), so all of the signs are switched), then the function is odd.

What is the inverse of a one-to-one function?

Theorem If f is a one-to-one continuous function defined on an interval, then its inverse f−1 is also one-to-one and continuous. (Thus f−1(x) has an inverse, which has to be f(x), by the equivalence of equations given in the definition of the inverse function.)