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

[demonelite123]

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.

 

[Defennder]

It isn't a one-to-one function to begin with, not unless you specify the function's domain.

 

[demonelite123]

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.

 

[Defennder]

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.

 

[demonelite123]

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.

 

[HallsofIvy]

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?

 

[demonelite123]

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.

 

[Defennder]

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?

 

[demonelite123]

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?

 

[HallsofIvy]

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.

 

[roam]

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 $$x_{1}^3 - 4x_{1}^2 + 2 = x_{2}^3 - 4x_{2}^2 + 2$$ is correct.

The so-called horizontal line test is a geometrical interpetation of what one to one means.

 

[HallsofIvy]

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.