Is |x^3-1| One to One? Monotonicity and Inverse Function Analysis

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In summary, the conversation discusses the function |x^3-1| and whether it is one-to-one and monotonous. It is determined that the function is not one-to-one and does not have an inverse function. Counterexamples are needed to prove this, which is further elaborated on in the conversation.
  • #1
peripatein
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Homework Statement



Is the function |x^3-1| one to one? Is it monotonous?

Homework Equations





The Attempt at a Solution



Since |x^3-1|=|y^3-1| does not necessarily imply that x=y for every x and y, I presume it is not one to one. Hence it has no inverse function.
It is also not monotonous.
Are all these statements correct?
 
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  • #2
yes :smile:
 
  • #3
Hi tiny-tim,
Once again, thanks a lot! :-)
 
  • #4
While the statements are correct, they are not sufficient to complete the exercise. To actually solve it, you need to find counterexamples. For example, if you want to show that [itex]|x^3-1|[/itex] is not one-to-one, you need to come up with two particular and distinct points x and y such that [itex]|x^3-1|=|y^3-1|[/itex]. Just saying that it is one-to-one is not enough without counterexample.
 
  • #5
Have done so, simply didn't specify it :-). Thank you, micromass!
 

FAQ: Is |x^3-1| One to One? Monotonicity and Inverse Function Analysis

What does it mean for a function to be one-to-one?

A one-to-one function is a function in which each element in the domain is mapped to a unique element in the range. This means that no two elements in the domain can have the same output.

How can I determine if |x^3-1| is a one-to-one function?

To determine if |x^3-1| is a one-to-one function, you can use the horizontal line test. If a horizontal line intersects the graph of the function at more than one point, then the function is not one-to-one. If a horizontal line intersects the graph at only one point, then the function is one-to-one.

What is the domain and range of |x^3-1|?

The domain of |x^3-1| is all real numbers, and the range is all non-negative real numbers (including 0).

Can |x^3-1| be both even and odd?

No, |x^3-1| cannot be both even and odd. A function can only be even if f(-x) = f(x) for all values of x, and it can only be odd if f(-x) = -f(x) for all values of x. However, |x^3-1| does not satisfy either of these conditions, so it is neither even nor odd.

How can I prove that |x^3-1| is a one-to-one function?

You can prove that |x^3-1| is a one-to-one function by showing that no two values in the domain have the same output. This can be done algebraically by setting the function equal to a constant and solving for x. If you get more than one solution, then the function is not one-to-one. If you get only one solution, then the function is one-to-one.

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