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Rolles Theorem, showing two distinct points.

  1. Feb 25, 2012 #1
    Have the following question and just wondering if my solution is correct

    Let g(x)= x^5+3x-1. Show that there are no distinct points x_1, x_2 in R such that g(x_1)=g(x_2).

    Proof by contradiction. Assume we have two solution x_1<x_2 in ℝ, i,e g(x_1)+g(x_2)=0, since g is differentiable on (x_1,x_2) and continuous on [x_1,x_2], then we can apply rolles theorem, there exits a C belonging to (x_1, x_2) such that df/dx=0, however df/dx=5x^4+3>0 Hence we have a contradiction and only one solution to f(x)=0.

    Many thanks in advance.
     
    Last edited by a moderator: Feb 25, 2012
  2. jcsd
  3. Feb 25, 2012 #2

    micromass

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    Sounds correct!!
     
  4. Feb 25, 2012 #3

    HallsofIvy

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    I presume you mean g(x_1)- g(x_2)= 0, not the sum.

     
  5. Feb 25, 2012 #4

    Mark44

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    Or you could just take the direct approach. Since g'(x) = 5x4 + 3, we see that g'(x) > 0 for all x, which means that the graph of g is increasing on the entire real line, hence g is one-to-one. This fact implies that if g(x1) = g(x2), then x1 = x2.
     
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