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Math Help - Proof

  1. #1
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    Proof

    If x is real , prove that the expression \frac{(x - a)(x - c)}{(x -b)} is capable of assuming all values provided that a , b , c are either in ascending or descending orders of magnitude .

    I'm lost.......how do I show that \frac{(x - a)(x - c)}{(x -b)} is "capable of assuming all values".
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  2. #2
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    You want to show that the range of this function is the set of reals. Showing it's one to one will suffice in this case since ( I am assuming)domain=co-domain=real set
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  3. #3
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    But how do I do that ?
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  4. #4
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    Hmm struggling myself. Obviously need to show f(x1)=f(x2) implies x1=x2 if you want to try.
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  5. #5
    A Plied Mathematician
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    Quote Originally Posted by boromir View Post
    You want to show that the range of this function is the set of reals.
    Right.

    Showing it's one to one will suffice in this case since ( I am assuming)domain=co-domain=real set
    But the function is not one-to-one, nor is the domain the set of all real numbers (x = b is not in the domain).

    Try setting an arbitrary

    y=\frac{(x-a)(x-c)}{x-b},

    and see if you can find a real x that satisfies the equation, no matter what y is.
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  6. #6
    Junior Member qspeechc's Avatar
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    Here are my thoughts on this.

    Let us call f(x) = \frac{(x - a)(x - c)}{(x -b)}

    Assume that a < b < c. For the other case you can try it for yourself, it should be very similar. Try sketching the situation I describe below.

    When x < a then each of the expressions (x-a), (x-b), (x-c) is negative, so f(x) is negative. As x\rightarrow -\infty (meaning as x becomes a larger and larger negative number) then f(x)\rightarrow -\infty (meaning the value of f(x) becomes a larger and larger negative number). Try explaining this for yourself.

    At x=a, f(a)=0.

    When a < x < b then f(x) is positive, which you can see by looking at whether each factor (x-a), (x-b), (x-c) is positive or negative. As x approaches b from below, the value of f(x) should approach \infty, in other words, f(x) should get larger and larger without bound (vertical asymptote).

    This already shows that f(x) assumes all values, so you don't have to look at what goes on elsewhere (i.e. x>b), but if you're interested you can check it out yourself.
    Well actually the missing piece is that the function is continuous on its domain, which simply means that there are no "breaks" in the graph.
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  7. #7
    Junior Member Renji Rodrigo's Avatar
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    consider the function  f:R\setminus\{b\} \rightarrow R


    with f(x) = \frac{(x-a)(x-c)}{(x-b)}. suppose a<b<c

    Consider the function on the interval (b,\infty ),

    we take the limit x \rightarrow b^{+}


    \lim_{x \rightarrow b^{+}}\frac{(x-a)(x-c)}{(x-b)}= -\infty


    because (b-a)>0 and b-c<0
    in the same way


    \lim_{x \rightarrow \infty}\frac{(x-a)(x-c)}{(x-b)}= \infty

    f is continuos so f assume all values on R in the interval (b,\infty )
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  8. #8
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    Rodrigo, I thought the intermediate value theorem stipulates that f has to be continuous on a closed interval for it to assume all values in it. Great work btw.
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  9. #9
    Junior Member Renji Rodrigo's Avatar
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    : )
    Boromir it's a corollary of the intermediate value theorem (IVT)

    If  f is continue in (a, b)=I with \lim_{x \rightarrow a^+} f(x) = - \infty , \lim_{x \rightarrow b^-} f(x) =  \infty so f(I)=R.



    Take any y  \in R , we will show that exist x \in (a, b) with f(x)=y .

    \lim_{x \rightarrow a^+} f(x) = - \infty , \lim_{x \rightarrow b^-} f(x) =  \infty

    so for any A, B positive , exists \delta_1 , \delta_2 positive with x \in  (a, a+ \delta_1 )=I_1 \Rightarrow  f(x) < -A,

    x \in  ( b- \delta_1, b )=I_2 \Rightarrow f(x) > B, we can take A and B with y \in (-A , B) ,

    exist \underbrace{x_1}_{\in I_1}<\underbrace{x_2}_{\in I_2} with

    f(x_1)<-A , f(x_2)>B , f(x_1)<y<f(x_2) then
    exist in [x_1,x_2] a number x_3 with f(x_3)= y by the IVT .

    If the interval is (a, \infty ) with \lim_{x \rightarrow \infty} f(x) =  \infty so for all B_2>0 exists B_1>0


    with x>B_1 and f(x)>B_2. Taking B_2 >y exist x_2> B_1, with f(x_2)>B_2 >y we use the same argument that we used before in the interval [x_1, x_2 ] .
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  10. #10
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    Must say my friend (mrfour44) gave me the idea with the discriminant. I see it's probably along the lines of what Ackbeet proposed. Just consider this method because it's easier (i.e. more elementary).
    So let's say our function is equal to an arbitrary real number, we have to prove that is possible. But by setting it equal to something we make it an equation and just have to show that it has a solution for real x.
    So, let a,b,c,R \in  R, a,b,c are in an ascending or descending order (i.e. a>b>c or a<b<c) and:

    \frac{(x-a)(x-c)}{x-b}=R

    Multiply by denominator and then just treat it like a quadratic:

    x^2-(a+c)x+ac=Rx-Rb
    x^2-(a+c+R)x+ac+Rb=0

    Now, I guess you probably know what the discriminant is. So we have said that this equation holds for some real x. Now we must prove that, and this amounts to proving that the discriminant is greater than or equal to zero. So we have:

    (a+c+R)^2-4(ac+Rb)\geqslant 0
    (a+c)^2+2R(a+c)+R^2-4ac-4Rb\geqslant 0
    a^2+2ac-4ac+c^2+2R(a+c)+R^2\geqslant 4Rb
    (a-c)^2+2R(a+c)+R^2\geqslant 4Rb

    Now, use the AG-GM inequality ( a+b\geqslant 2\sqrt{ab}) on (a-c)^2 and R^2:

    (a-c)^2+R^2\geqslant2\sqrt{(a-c)^2R^2}=2|a-c||R|

    And combine with the initial inequality:

    (a-c)^2+R^2+2R(a+c)\geqslant 2|a-c||R|+2R(a+c)\geqslant4Rb

    The rest of it is just checking whether this weaker ineqaulity is true for all the possibilities and those are (obviously it's true for R=0):

    1. a>b>c and R>0
    2. a>b>c and R<0
    3. a<b<c and R>0
    4. a<b<c and R<0

    Now I won't go into that. This proof is long anyway. But this part is (considering this whole problem) rather trivial.
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