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Silly Question about systems of equations

  1. Sep 29, 2007 #1
    I had a thought while doing some calculus last night and it now concerns me that for some reason I cannot answer it. If given the equations [tex]y=x[/tex] and [tex]y=x^2+1[/tex] it would appear that since the range of the latter "reaches" infinity "first" that these two curves never meet. Now that thought did not occur to me until I sketched them after I tried solving them as a system of equations.

    Is there some way (not intuitively, but abstractly) to see that these two curves never intersect?

  2. jcsd
  3. Sep 29, 2007 #2

    D H

    Staff: Mentor

    These two equations specifically, sure. x2+1=x obviously has no solutions in the reals.

    Any system of equations, obviously not. There is no general solution for finding the zeros of a quintic, let alone any arbitrary function.
  4. Sep 29, 2007 #3
    I am not interested in finding any zeros; I am interested in knowing if there is a way to determine that there are not any zeros.

  5. Sep 29, 2007 #4


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    Staff Emeritus
    Science Advisor

    Look at f(x)= x2+1- x. f'(x)= 2x- 1 which is positive for x> 1/2 so it is always increasing. When x= 1/2, that has a value of 1/4+ 1- 1/2= 3/4. As long as there is no zero between 0 and 1/2, the function cannot have a zero and so the equation
    x2+ 1= x cannot have a solution.
    Last edited by a moderator: Oct 2, 2007
  6. Sep 29, 2007 #5

    D H

    Staff: Mentor

    There is no general method to determine whether some function [itex]f(x)[/itex] has any zeroes. Suppose you evaluate the function at two points, [itex]x_1[/itex] and [itex]x_2[/itex]. If the function value is positive at both points (or negative at both points), there might well be no points or two points or four points (or ...) between [itex]x_1[/itex] and [itex]x_2[/itex] where the function is zero. (Note: In saying this I am counting double zeros such as x2=0 as two zeros). You don't know which is the case and there is no general way of knowing.

    Suppose you get lucky and find that [itex]f(x_1)[/itex] and [itex]f(x_2)[/itex] have opposite signs. You still don't know that the function passes through zero between [itex]x_1[/itex] and [itex]x_2[/itex]. Consider [itex]f(x)=1/x[/itex] for example. [itex]f(-2)=-1/2[/itex] and [itex]f(1)=1[/itex], but the function does not pass through zero between -2 and 1. In the special case tht [itex]f(x)[/tex] is continuous and exists everywhere between [itex]x_1[/itex] and [itex]x_2[/itex] then finding points with opposite signs does guarantee that the function is zero at at least one point between [itex]x_1[/itex] and [itex]x_2[/itex].
  7. Sep 29, 2007 #6
    Gotcha. Thanks for explaining D H.

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