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How do u prove that this is a one to one function 
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#1
Aug3108, 12:39 AM

P: 219

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. 


#2
Aug3108, 01:16 AM

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P: 2,616

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



#3
Aug3108, 01:29 AM

P: 219




#4
Aug3108, 01:54 AM

HW Helper
P: 2,616

How do u prove that this is a one to one function
I really don't see how your method works to show that it is not onetoone by contradiction. Suppose that it is indeed onetoone, 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
Aug3108, 02:18 AM

P: 219




#6
Aug3108, 04:25 AM

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


#7
Sep108, 09:02 PM

P: 219

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
Sep108, 10:00 PM

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Well, what does that tell you about how many times the graph crosses the xaxis? And what does that in turn tell you about whether it's oneone?



#9
Sep208, 08:54 PM

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#10
Sep308, 03:31 AM

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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
Sep308, 04:53 AM

P: 895

To show that y is one to one, it is required to show that if y(x_{1}) = y(x_{2}), then x_{1} = x_{2}
I think your [tex]x_{1}^3  4x_{1}^2 + 2 = x_{2}^3  4x_{2}^2 + 2[/tex] is correct. The socalled horizontal line test is a geometrical interpetation of what one to one means. 


#12
Sep308, 09:58 AM

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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.



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