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Math Help - Mean value theorem & abs maximum/minimum calculus?

  1. #1
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    Mean value theorem & abs maximum/minimum calculus?

    a) Find all points in the interval [-1, 2] where the function f(x) = x^3 - x^2 satisfi es


    the conclusion of the Mean Value Theorem.

    I don't get it...i get to the point where i get 3x^2 - 2x = 2 , then i don't know how to get x

    b) f(x) = ln(x + 1) with domain [0; 3].


    i got abs maximum as f(3) = ln4 and abs minimum as f(0)=ln1 ..lemme know if that is correct.


    c)

    Prove that the function f(x) = x^3 + 7 + 4= x^2 has at most one zero in the interval
    (-inf, 0).

    thanks
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  2. #2
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    Re: Mean value theorem & abs maximum/minimum calculus?

    Quote Originally Posted by qwerty999 View Post
    a) Find all points in the interval [-1, 2] where the function f(x) = x^3 - x^2 satisfies


    the conclusion of the Mean Value Theorem.

    I don't get it...i get to the point where i get 3x^2 - 2x = 2 , then i don't know how to get x

    3x^2 - 2x - 2 = 0 , use the quadratic formula

    b) f(x) = ln(x + 1) with domain [0; 3].


    i got abs maximum as f(3) = ln4 and abs minimum as f(0)=ln1 ..lemme know if that is correct.


    why don't you sketch the graph of the function over the given interval and check it yourself?

    c)

    Find all points in the interval [-1, 2] where the function f(x) = x^3 - x^2 satisfies the conclusion of the Mean Value Theorem.

    thanks


    (a) and (c) are the same problem
    ...
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  3. #3
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    Re: Mean value theorem & abs maximum/minimum calculus?

    @skeeter ..so are u saying that my answer for a) is correct so far and i have to use the quadratic formula now for the final answer. For b) I don't
    know how to do that. I changed c)
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  4. #4
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    Re: Mean value theorem & abs maximum/minimum calculus?

    Quote Originally Posted by qwerty999 View Post
    @skeeter ..so are u saying that my answer for a) is correct so far and i have to use the quadratic formula now for the final answer. For b) I don't
    know how to do that. I changed c)
    (a) correct so far

    (b) you're telling me that you cannot sketch the graph of a very basic function like y = ln(x+1) over the given domain?

    (c)
    f(x) = x^3 + 7 + 4= x^2
    this makes no sense ... check it again.
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