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One-to-one'ess question

  • Thread starter mindauggas
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  • #1
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Homework Statement



Is there a way to show that the function [itex]\frac{2x^{2}}{x-2}[/itex] is/is'nt one to one, non graphicaly (without drawing the graph)?

The Attempt at a Solution



I tried to following: To determine whether this function is one to one, look at what happens if two values of x give the same y: suppose [itex]2a^{2}/(a-2)= 2b^{2}/(b-2)[/itex]. Multιply both sides by [itex](a-2)(b-2)[/itex] to get [itex]2a^{2}(b-2)= 2b^{2}(a-2)[/itex]. That is the same as [itex]2a^{2}b-2a= 2b^{2}a-2b[/itex]. Trying to show, that that two different values of x cannot give the same y (this is an adaptation sugested by HallsofIvy on a similar problem). But got stuck.

Help much needed here :)
 
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Answers and Replies

  • #2
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Isn't it easier to try and find an inverse function?
 
  • #3
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Isn't it easier to try and find an inverse function?
Does the inability to find an inverse imply the absence of one-to-one'ess whit strict necesity?
On the other hand how will i know that the inverse does not exist (instead of not being able to find it) ??

Are there any analitical methods to determine one-to-one'ess in general?
 
  • #4
LCKurtz
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Homework Statement



Is there a way to show that the function [itex]\frac{2x^{2}}{x-2}[/itex] is/is'nt one to one, non graphicaly (without drawing the graph)?

The Attempt at a Solution



I tried to following: To determine whether this function is one to one, look at what happens if two values of x give the same y: suppose [itex]2a^{2}/(a-2)= 2b^{2}/(b-2)[/itex]. Multιply both sides by [itex](a-2)(b-2)[/itex] to get [itex]2a^{2}(b-2)= 2b^{2}(a-2)[/itex]. That is the same as [itex]2a^{2}b-2a= 2b^{2}a-2b[/itex]. Trying to show, that that two different values of x cannot give the same y (this is an adaptation sugested by HallsofIvy on a similar problem). But got stuck.

Help much needed here :)
Just keep working:
##a^2(b-2)=b^2(a-2)##
##a^2b-2a^2-ab^2+2b^2=0##
##ab(a-b)-2(a+b)(a-b)=0##
Since we don't want ##a=b## divide out the ##a-b##.
##ab-2a-2b=0##
##a(b-2)=2b##
##a = \frac{2b}{b-2}## Take, for example, b = 3, a = 6.
 
  • #5
SammyS
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Does the inability to find an inverse imply the absence of one-to-one'ness with strict necessity?
...
No, but if solving y=f(x) for x gives a result with a ± sign, then you know the function is not one-to-one.

Some other ways you might know that this function is not one-to-one are at least somewhat related to the graph y=f(x), even though you can determine them without actually graphing the function.

The graph has a slant asymptote of y = x + 4, and a vertical asymptote at x=2. These two thing together are incompatible with being one-to-one.

The function has a zero of even multiplicity (multiplicity 2) thus it is not one-to-one .
 
  • #6
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The graph has a slant asymptote of y = x + 4, and a vertical asymptote at x=2. These two thing together are incompatible with being one-to-one.
Thanks for these criteria of determining not-one-to-one'ness. But could you elaborate on this quoted part? why are these asymptes incompatible with a function whitch has them being one-to-one?
 
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  • #7
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Just keep working:
##a^2(b-2)=b^2(a-2)##
##a^2b-2a^2-ab^2+2b^2=0##
##ab(a-b)-2(a+b)(a-b)=0##
Since we don't want ##a=b## divide out the ##a-b##.
##ab-2a-2b=0##
##a(b-2)=2b##
##a = \frac{2b}{b-2}## Take, for example, b = 3, a = 6.
Thanks :)
 
  • #8
LCKurtz
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No, but if solving y=f(x) for x gives a result with a ± sign, then you know the function is not one-to-one.

Some other ways you might know that this function is not one-to-one...

The function has a zero of even multiplicity (multiplicity 2) thus it is not one-to-one .
I'm not so sure about that one. Consider$$
f(x) =\left \{ \begin{array}{rl}
x^3,& x\le 0\\
x^2,&x>0
\end{array}\right.$$Here both f(0) and f'(0) are 0 and f''(0) doesn't exist, so it has a root of multiplicity 2.

[Edit, added later:] Looking at a couple of alternate definitions of multiplicity on the internet, f''(0) not existing might kill this counterexample. This may not be multiplicity 2. Also the definition in terms of limit is that a root r is multiplicity n if$$
\lim_{x\rightarrow r}\frac{f(x)}{(x-r)^n}$$is a non-zero number. That doesn't work for my example either.
 
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  • #9
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Just keep working:
##a^2(b-2)=b^2(a-2)##
##a^2b-2a^2-ab^2+2b^2=0##
##ab(a-b)-2(a+b)(a-b)=0##
Since we don't want ##a=b## divide out the ##a-b##.
##ab-2a-2b=0##
##a(b-2)=2b##
##a = \frac{2b}{b-2}## Take, for example, b = 3, a = 6.
Sorry, for returning to this exercise after such a long time, but I realise that I have doubts about how to interpret the result. If we do not get a=b, that means that there are more than one x in terms of which y can be expressed? Thus making the function not one-to-one?
 
  • #10
LCKurtz
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Sorry, for returning to this exercise after such a long time, but I realise that I have doubts about how to interpret the result. If we do not get a=b, that means that there are more than one x in terms of which y can be expressed? Thus making the function not one-to-one?
You were looking for two different values ##x=a## and ##x=b## such such that$$
f(x)=\frac {2x^2}{x-2}$$had the same value. My example in post #4 was ##x=3## and ##x=6##. Do they work?
 
  • #11
HallsofIvy
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In your first post, you started by looking at
[tex]\frac{2a^2}{a- 2}= \frac{2b^2}{b- 2}[/tex]
and multiplied both sides by (a- 2)(b- 2) to get [itex]2a^2(b- 2)= 2b^2(a- 2)[/itex]]

As LCKurtz pointed out: You can continue that to get [itex]a^2b- 2a^2= ab^2- 2b^2[/itex] and you can write that as [itex]a^2b- ab^2= 2(a^2- b^2)[/itex], [itex] (a- b)(ab)= 2(a- b)(a+ b)[/itex]. With [itex]a\ne b[/itex], we can divide by a- b to get [itex]ab= 2(a+ b)[/itex]. What LCKurtz then did was choose values of a and b that satisfy that: if a= 3 and b= 6, ab= (3)(6)= 18= 2(9)= 2(3+ 6).
 
  • #12
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You were looking for two different values ##x=a## and ##x=b## such such that$$
f(x)=\frac {2x^2}{x-2}$$had the same value. My example in post #4 was ##x=3## and ##x=6##. Do they work?
Yes, they do give the same value.
 
  • #13
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In your first post, you started by looking at
[tex]\frac{2a^2}{a- 2}= \frac{2b^2}{b- 2}[/tex]
and multiplied both sides by (a- 2)(b- 2) to get [itex]2a^2(b- 2)= 2b^2(a- 2)[/itex]]

As LCKurtz pointed out: You can continue that to get [itex]a^2b- 2a^2= ab^2- 2b^2[/itex] and you can write that as [itex]a^2b- ab^2= 2(a^2- b^2)[/itex], [itex] (a- b)(ab)= 2(a- b)(a+ b)[/itex]. With [itex]a\ne b[/itex], we can divide by a- b to get [itex]ab= 2(a+ b)[/itex]. What LCKurtz then did was choose values of a and b that satisfy that: if a= 3 and b= 6, ab= (3)(6)= 18= 2(9)= 2(3+ 6).
i understand what was done and sugested, but what does it mean exactly and precisely in terms of one-to-ones? For example: "this the function is nor one to one, because we got two values of x in that give the same y". But what then is the meaning of the derived term a=2b/b-2 ? Is it just a tool to generate posible values of x for which we would get the same y? A mental aid? Or does it necesarily give all the x's for which we get the same y? (for example b=1 gives a=-2 and they both give the same f(x) when put into the equation).
 
  • #14
LCKurtz
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Homework Statement



Is there a way to show that the function [itex]\frac{2x^{2}}{x-2}[/itex] is/is'nt one to one, non graphicaly (without drawing the graph)?

The Attempt at a Solution



I tried to following: To determine whether this function is one to one, look at what happens if two values of x give the same y: suppose [itex]2a^{2}/(a-2)= 2b^{2}/(b-2)[/itex]. Multιply both sides by [itex](a-2)(b-2)[/itex] to get [itex]2a^{2}(b-2)= 2b^{2}(a-2)[/itex]. That is the same as [itex]2a^{2}b-2a= 2b^{2}a-2b[/itex].
i understand what was done and sugested, but what does it mean exactly and precisely in terms of one-to-ones? For example: "this the function is nor one to one, because we got two values of x in that give the same y". But what then is the meaning of the derived term a=2b/b-2 ? Is it just a tool to generate posible values of x for which we would get the same y? A mental aid? Or does it necesarily give all the x's for which we get the same y? (for example b=1 gives a=-2 and they both give the same f(x) when put into the equation).
You started by setting f(a) = f(b) so you had two, hopefully different, x's giving the same y. The idea was to see if that was possible or not because that will tell you whether f is 1-1. The end result of that was the equation a = 2b/(b-2). In particular, it did not come that a had to equal b. All you have to do to show the function isn't 1-1 is to find 2 values of x giving the same f(x). b = 3 and a = 6 are one choice. b = 1 an a = -2 are others. There are more, but you only need one.
 
  • #15
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Clear enought. Thanks :)
 
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