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Math Help - Is there a number that is exactly 1 more than its cube?

  1. #1
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    Is there a number that is exactly 1 more than its cube?

    So, Im studying limits, and got to the incredible cool intermediate value theorem now, and found this question in my book. Even though he doesnt ask for THE number, he wants me to show that it exists, so i tried:

    x = x + 1

    now, here is where the problem stats, i can form two different functions,
    f(x) =  x + 1 - x
    g(x) = -x - 1 + x

    By testing, however, the answer is the same, x is in (-1,4 , -1,3). But when I actually look at the results, I notice that:
    f(-1,3) = 0,103
    g(-1,3) = -0,103

    f(-1,4) = -0,344
    g(-1,4) = 0,344

    Even though in this case the function I choose wont matter, since I will get to the answer, Should I have picked one over another?
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  2. #2
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    Quote Originally Posted by Capes View Post
    So, Im studying limits, and got to the incredible cool intermediate value theorem now, and found this question in my book. Even though he doesnt ask for THE number, he wants me to show that it exists, so i tried:

    x = x + 1
    f(x) =  x + 1 - x
    Even though in this case the function I choose wont matter, since I will get to the answer, Should I have picked one over another?
    No, just note that f(-2)<0<f(1).
    That is sufficient to show that f(x)=0 for some x.
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  3. #3
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    For your function \displaystyle y = x^3 - x + 1, note that when \displaystyle x = -2, y = -5, and when \displaystyle x = 1, y = 1.

    Since the function's value has gone from something negative to something positive, it must have crossed the \displaystyle x axis somewhere at least once.

    So a solution exists.


    Edit: Crap, just realised that Plato posted the exact same thing - serves me right for not reading the responses more carefully haha.
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