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R0lle's Theorem

  1. Aug 26, 2008 #1
    Use Rolle's theorem to show that [tex]g(x) = 3x^4 +4x^3 +6x^2 -3[/tex] has exactly two real roots.



    3. The attempt at a solution:

    Rolle's theorem states that if g is continious on the closed interval [a,b] and differentiable on the open interval (a,b). Assuming that g(a) = g(b). Then g'(x0) = 0 for at least one point x0 in (a,b).

    Looking at the derivative of that function:
    [tex]g'(x) = 12x^3 + 12x^2 + 12x[/tex]

    the function is equal to zero when;
    [tex]12x^3 + 12x^2 + 12x = 0[/tex]
    [tex]x (x^2 + x + 1) = 0[/tex]

    I used the quadratic formula and I believe x^2 + x + 1 has no real roots which means x = 0 is the only real root for the derivative function. Thus x = 0 is the only "stationary point" on the curve for the original function g(x).

    I only found one stationary point! How does this show that "there can be at most 2 real roots"?

    How can I use Rolle's theorem to prove that g has at most two real roots?

    (I believe at x = 0 the function is negative and it perhaps crosses the x-axis but I don't know how to find the coordinates of the roots!)


     
  2. jcsd
  3. Aug 26, 2008 #2

    tiny-tim

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    higher-order version?



    Hi roam! :smile:

    :confused: the only thing I can think of is that perhaps they mean the higher-order version of Rolle's theorem …

    If g(a) = g(b) = g(c) with a < b < c, then there must be an x with a < x < c and g''(x) = 0.
     
  4. Aug 26, 2008 #3

    HallsofIvy

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    I don't know if you have really "used" Rolles theorem but it is true that y'= 0 only at x= 0. Now, look at x= 1. y'(1)= 1(12+ 1+ 1)= 3> 0. Since y'= 0 only at 0 (and y' is continuous) that tells that y'> 0 and y is increasing for all x> 0. Look at x= -1. y'(-1)= (-1)((-1)2+ (-1)+ 1)= (-1)(1)= -1< 0. That tells you that y'< 0 and y is decreasing for all x< 0. That should be enough to tell you that y cannot have more than rwo zeros.
     
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