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Math Help - Calc Question

  1. #1
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    Calc Question

    Let f be the function given by f(x)= (x^3)-(6x^2)+p, where p is an arbitrayry constant.

    a)Write an expression for f'(x) and use it to find the relative maximum and minimum values of f in terms of p. Show the analysis that leads to your conclusion.

    b)For what values of the constant p does f have 3 distinct roots?

    c)Find the value of p such that the average value of f over the closed interval [-1,2] is 1.


    For a), I found the derivative, which is f'(x)=(3x^2)-12x. I set that equal to 0, which gives me x=0,4. I put these values back into the original equation and got f(0)=o and f(4)=-32. In terms of p, I got 0=p and 32=p. Is this right?

    I have no clue how to do b). I kind of know how to do c) but I would appreciate some help with it. Thanks a lot.
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  2. #2
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    Quote Originally Posted by Vigo
    Let f be the function given by f(x)= (x^3)-(6x^2)+p, where p is an arbitrayry constant.

    a)Write an expression for f'(x) and use it to find the relative maximum and minimum values of f in terms of p. Show the analysis that leads to your conclusion.
    f(x)=x^3-6x^2+p
    f'(x)=3x^2-12x
    f''(x)=6x-12
    Find the critical points, f'(x)=0 thus,
    3x^2-12x=0
    x(3x-12)=0
    x=4,0
    But, f''(0)<0 \mbox{ and }f''(4)>0
    Thus, x=0 is maximum (2nd derivative test)
    and x=4 is minimum.
    But, f(0)=p and f(4)=-32+p
    Thus, min at (4,-32+p) max at (0,p)
    \mathbb{Q}.\mathbb{E}.\mathbb{D}
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