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Math Help - Calculus ProofS Help

  1. #1
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    Calculus ProofS Help

    Hey, I need help on a couple of problems in my proof class. If anyone can help, it would be highly appreciated, Thank you!

    1. Prove that S = { n−1 |n ∈ N} is bounded above and that its supremumn
    is equal to 1.
    2.Use the Intermediate Value Theorem to show that the polynomial x4 +
    x3 − 9 has at least two real roots.
    3.Use the Mean Value Theorem to prove that |sin(b) − sin(a)| ≤ |b − a|
    for all a, b ∈ R.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Swamifez View Post
    2.Use the Intermediate Value Theorem to show that the polynomial x4 +
    x3 − 9 has at least two real roots.
    Let f(x) = x^4+x^3-9.

    Note that for large, negative x f(x) takes a positive value.
    f(0)= -9 is negative. Thus there is at least one root between a large negative x and x = 0.
    For large, positive x f(x) takes a positive value again. Thus there is at least one root between x = 0 and large a large positive x.

    Thus there are at least 2 real roots.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Swamifez View Post
    1. Prove that S = { n−1 |n ∈ N} is bounded above and that its supremumn
    is equal to 1.
    N = {1, 2, 3,...} yes? Then S = {0, 1, 2, ...}. This set is NOT bounded above!

    -Dan

    Ah wait! Are you defining S = \{ n^{-1} | n \in N \}?

    Then show that for p < q p^{-1} > q^{-1} ( p, q \in N). So the largest value in S is going to be for the smallest value in N: n = 1. Thus max(S) = 1. Since S has a maximum element sup(S) = max(S) (I believe. Someone please correct me if I'm wrong here!)

    -Dan
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  4. #4
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    Quote Originally Posted by Swamifez View Post
    3.Use the Mean Value Theorem to prove that |sin(b) − sin(a)| ≤ |b − a|
    for all a, b ∈ R.
    If b=a then the result is trivial (I love saying that).

    If b\not = a then without loss of generality (I love saying that too) you can assume b>a
    Then, this [a,b] is a closed interval.
    Consider the function, f(x)=\sin x defined on this interval. The function is continous on [a,b] and differenciable on (a,b) this satisfies the conditions of Lagrange's Mean Value Theorem. Thus, there exists a c\in (a,b) such as,
    \frac{\sin b-\sin a}{b-a}=f'(c)=cos c
    Thus,
    \left| \frac{\sin b-\sin a}{b-a} \right|=|\cos c|\leq 1
    Thus,
    |\sin b-\sin a|\leq |b-a|
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