How do u prove that this is a one to one function

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Any function that crosses its inverse is not one to one. And you can see that it crosses its inverse in 3 places. Yes, the "horizontal line test" is a geometric interpretation of "one to one". If a horizontal line cuts the graph of the function in more than one place, it is not one to one. If it doesn't, it is. That is a very general rule, not specific to any particular type of function.In summary, the given function y = x^3 - 4x^2 + 2 is not one to one. This can be shown algebraically by setting f(a) = f(b) and solving for a and b, which leads to the contradiction
  • #1
demonelite123
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how do you prove that this is a one to one function algebraically?
y = x^3 - 4x^2 + 2

this is what I've done so far:
f(a) = f(b), a=/=b

a^3 - 4a^2 +2 = b^3 - 4b^2 +2
a^3 - 4a^2 = b^3 - 4b^2 (subtract 2 from both sides)
a^3 - b^3 - 4a^2 + 4b^2 = 0
(a - b)(a^2 + ab + b^2) - 4(a^2 - b^2) = 0
(a - b)(a^2 + ab + b^2) - 4(a + b)(a - b) = 0
(a - b)(a^2 + ab + b^2 - 4a - 4b) = 0

i have no idea what to do after this. i know there are probably easier ways of determining whether a function is one to one or not but my teacher wants us to do it this way.
 
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  • #2
It isn't a one-to-one function to begin with, not unless you specify the function's domain.
 
  • #3
Defennder said:
It isn't a one-to-one function to begin with, not unless you specify the function's domain.

well all that you are given is that function and you have to determine if it is one to one or not. using the zero property i know that from the last step i left off from one of the solutions is a = b. now the other factor is (a^2 + ab + b^2 - 4a - 4b) = 0. i just want to know how to solve that portion if possible and find the solution so i can see whether the two solutions contradict the given statements f(a) = f(b), a=/=b or not.
 
  • #4
I really don't see how your method works to show that it is not one-to-one by contradiction. Suppose that it is indeed one-to-one, then a=b and your equation says 0=0. What can you deduce?

On the other hand, doing it algebraically means you have to solve some cubic equation by Cardano's method. I don't think that the problem is that complicated.
 
  • #5
Defennder said:
I really don't see how your method works to show that it is not one-to-one by contradiction. Suppose that it is indeed one-to-one, then a=b and your equation says 0=0. What can you deduce?

On the other hand, doing it algebraically means you have to solve some cubic equation by Cardano's method. I don't think that the problem is that complicated.

can you show me how you would show that this function is one to one algebraically? besides graphing the function, this is the only method i was taught for determining whether functions are one to one or not. i have been using this method for all of my homework problems and this is the only problem where it seemingly doesn't work very well.
 
  • #6
demonelite123 said:
well all that you are given is that function and you have to determine if it is one to one or not. using the zero property i know that from the last step i left off from one of the solutions is a = b. now the other factor is (a^2 + ab + b^2 - 4a - 4b) = 0. i just want to know how to solve that portion if possible and find the solution so i can see whether the two solutions contradict the given statements f(a) = f(b), a=/=b or not.
You initially asked how to prove it WAS one to one. Now you are saying "determine IF it is one to one or not". Those are very different!

What are f(-1), f(0), and f(1)? What do they tell you?
 
  • #7
HallsofIvy said:
You initially asked how to prove it WAS one to one. Now you are saying "determine IF it is one to one or not". Those are very different!

What are f(-1), f(0), and f(1)? What do they tell you?

well f(-1) = (-1)^3 - 4(-1)^2 + 2 = -3
f(0) = 0^3 - 4(0)^2 + 2 = 2
f(1) = 1^3 - 4(1)^2 + 2 = -1

i'm not sure what i should get from this.
 
  • #8
Well, what does that tell you about how many times the graph crosses the x-axis? And what does that in turn tell you about whether it's one-one?
 
  • #9
Defennder said:
Well, what does that tell you about how many times the graph crosses the x-axis? And what does that in turn tell you about whether it's one-one?

so it crosses the x-axis 3 times? did you just pick 3 random points or something?
 
  • #10
demonelite123 said:
so it crosses the x-axis 3 times? did you just pick 3 random points or something?
No, it doesn't cross the x-axis 3 times. The function value is -3 at x= -1 and 2 at x= 0. That means it equals 0 for some x between -1 and 0. The function value is 2 at x= 0 and -1 at x= 1. That means it equals 0 for some x between 0 and 1.

No, I didn't pick 3 random points. I graphed the function so I could see immediately whether it was one to one or not.
 
  • #11
To show that y is one to one, it is required to show that if y(x1) = y(x2), then x1 = x2

I think your [tex]x_{1}^3 - 4x_{1}^2 + 2 = x_{2}^3 - 4x_{2}^2 + 2[/tex] is correct.

The so-called horizontal line test is a geometrical interpetation of what one to one means.
 
  • #12
Yes, but the problem, as we were finally told, was NOT to "show that y is one to one". It was to determine WHETHER y= f(x) is one to one or not. It isn't.
 

1. How do you determine if a function is one-to-one?

To prove that a function is one-to-one, we need to show that for every element in the domain, there is a unique element in the range. This can be done by using the vertical line test or by showing that the function passes the horizontal line test.

2. What is the difference between one-to-one and onto functions?

A one-to-one function is a function where each element in the domain is mapped to a unique element in the range. On the other hand, an onto function is a function where every element in the range has at least one corresponding element in the domain. In other words, every element in the range is mapped to from the domain.

3. Can a one-to-one function have multiple outputs for the same input?

No, a one-to-one function cannot have multiple outputs for the same input. This would violate the definition of a one-to-one function, which states that each element in the domain must have a unique element in the range.

4. How do you prove that a function is not one-to-one?

To prove that a function is not one-to-one, we can use a counterexample. This means finding two distinct elements in the domain that are mapped to the same element in the range. This would show that the function is not one-to-one because there are multiple outputs for the same input.

5. Can a function be both one-to-one and onto?

Yes, a function can be both one-to-one and onto. This type of function is called a bijection. It means that each element in the domain is mapped to a unique element in the range, and every element in the range has at least one corresponding element in the domain. In other words, there is a one-to-one correspondence between the elements in the domain and the elements in the range.

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