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Thread: x^2-cosx-x

  1. #1
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    x^2-cosx-x

    [0,pi]
    i used ivt. But I said if is continuous on [0,pi]
    or should I have out f is a polynomial so continuous on [0,pi]?
    is this equation considered a polynomial I didn't know if it was not or is . I know a polynomial consists of several terms.
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  2. #2
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    Re: x^2-cosx-x

    Quote Originally Posted by math951 View Post
    [0,pi]
    i used ivt. But I said if is continuous on [0,pi]
    or should I have out f is a polynomial so continuous on [0,pi]?
    is this equation considered a polynomial I didn't know if it was not or is . I know a polynomial consists of several terms.
    what is the question/problem statement?
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  3. #3
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    Re: x^2-cosx-x

    Quote Originally Posted by math951 View Post
    [0,pi]
    i used ivt. But I said if is continuous on [0,pi]
    or should I have out f is a polynomial so continuous on [0,pi]?
    is this equation considered a polynomial I didn't know if it was not or is . I know a polynomial consists of several terms.
    No, this is definitely not a polynomial. A polynomial is several terms, yes, but those terms have to be a number times a non-negative integer power of x.

    Now, what is the question? Is it whether or not this function is continuous on 0 to pi? In that case you would need to know that powers of x are continuous for all x, that cosine is continuous for all x, and that sums of continuous functions are continuous. (So are differences, products and divisions as long as the divisor is non-zero.)
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    Re: x^2-cosx-x

    Is this equation an polynomial? Or is it not. Polynomials have infinity possible answers.
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    Re: x^2-cosx-x

    Infinite* cool ty for answer
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    Re: x^2-cosx-x

    Quote Originally Posted by math951 View Post
    Is this equation an polynomial? Or is it not. Polynomials have infinity possible answers.
    "Infinity possible answers" to what question? You do not "answer" a polynomial! If you mean "possible answers to a polynomial equation", this is still wrong. A polynomial equation, of degree n, has n solutions, counting multiple solutions and complex number solutions.

    (I responded, just as you were typing this, apparently, that "No, this is not a polynomial".)
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