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Homework Help: Real analysis question: show that x^4 - x - 1 = 0 has two real roots

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Let f(x)=x^4 - x - 1. Show that f(x)=0 has two real roots.

    2. Relevant equations


    3. The attempt at a solution

    x(x^3 - 1 - 1/x) = 0 which gives x=0 and x^3 - 1 - 1/x=0, x^2 - 1/x - 1/x^2=0, but WolframAlpha says x~~0.724492 and x~~-1.22074. I kept dividing by x it but couldn't come up with a sound result.
    Last edited: Nov 5, 2011
  2. jcsd
  3. Nov 5, 2011 #2


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    Homework Helper

    I don't think the point of this problem is to find the roots, but just to show they exist. Are you allowed to use the intermediate value theorem?
  4. Nov 5, 2011 #3


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    Find out the behaviour of the function as x goes either to infinity and to minus infinity. It is also useful to calculate f(0), f(1), f(-1), so as you get a picture of the function. Does it have minima and maxima? how many?

  5. Nov 8, 2011 #4
    Yes I am allowed to use the intermediate value theorem.

    as tends to inf: lim f(x) = inf
    as tends to -inf: lim f(x) = -inf

    I'm not sure on how to find the extremum though
  6. Nov 8, 2011 #5


    Staff: Mentor

    No, this isn't true. As x approaches -∞, f(x) approaches +∞.
    The problem is not asking for this. Focus on what the problem is asking you to show. It might be helpful to sketch a rough graph of this function based on the two limits and the three function values above. Keep in mind the suggestion from gb7nash.
    Last edited: Nov 8, 2011
  7. Nov 8, 2011 #6


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    prove that: if f(x) = x4 - x - 1,

    a) f tends to ∞ as x tends to -∞ or ∞.
    b) f'(x) has at least one root
    c) f has a global minimum
    d) this minimum is < 0.

    why does this tell you f has at least 2 real roots?

    then show f has at most 2 roots (hint: look at part (b) again).
  8. Nov 8, 2011 #7


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    You see that the function is positive at x=-1, then negative at x=0. As it is continuous, it must be zero somewhere between -1 and 0. It is -1 again at x=-1, so there is a turning point, a minimum between x=0 and x=1. And the function is positive for large x values. Is it zero again somewhere between x=1 and infinity?

  9. Nov 9, 2011 #8
    OK I figured this one out thanks to you guys:

    from the I.V. theorem, there exists c=0 belonging to [a, b] with a=1 and b=-1 in the first case, second case with a=1 and b=2. therefore f(a)=-1<c=0<f(b)=1 and f(a)=-1<c=0<f(2)=13 so we have two real values for x.

    Thanks y'all for your help!
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