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Math Help - Rolle's Theorem

  1. #1
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    Rolle's Theorem

    How do I do this problem?

    Use Rolle's theorem to prove that, regardless of the value b, there is at most one point in the interval -1 <= x <= 1 for which x^3 - 3x + b = 0
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  2. #2
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    Quote Originally Posted by BrianW View Post
    How do I do this problem?

    Use Rolle's theorem to prove that, regardless of the value b, there is at most one point in the interval -1 <= x <= 1 for which x^3 - 3x + b = 0
    We need to understand this problem carefully, because when I was first reading it, it seemed as if it said "there is at least one point which is a zero".

    We want to show that there is at most one point. We will argue by contradiction. Say there are more than two points on [-1,1]. Pick any two distinct ones x_1,x_2\in [-1,1] with x_1 < x_2. Define the function f(x) = x^3 - 3x + b. Note that f(x_1) = 0 \mbox{ and }f(x_2)=0. Consider this function f on the closed interval [x_1,x_2]. We notice that f(x_1) = f(x_2), the function is continous on [x_1,x_2], and differenciable on (x_1,x_2). By Rolle's theorem we can find x_0 \in (x_1,x_2) so that f'(x_0)=0. Meaning 3x_0^2 - 3 = 0. But that is impossible because 3x_0^2 - 3 < 0 since - 1< x_0 < 1. Hence a contradiction. That means there can be at most one zero.
    Q.E.D.
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