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Math Help - polynomial with max $ min

  1. #1
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    polynomial with max $ min

    Let p(x) be a real polynomials of Least Degree Which has Local Maximum at x=1 and Local Minimum


    at x=3.If p(1)=6 and p(3)=2. Then p^{'}(0)=
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  2. #2
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    Re: polynomial with max $ min

    Quote Originally Posted by jacks View Post
    Let p(x) be a real polynomials of Least Degree Which has Local Maximum at x=1 and Local Minimum


    at x=3.If p(1)=6 and p(3)=2. Then p^{'}(0)=
    p(x)=x^3-6x^2+9x+2 ~\text { so }~ p'(x)=3x^2-12x+9

    Hence :

    p'(0)=9
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  3. #3
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    Re: polynomial with max $ min

    Hello, jacks!

    \text{Let }p(x)\text{ be a polynomial of least degree which has local max at }x=1
    \text{and local minimum at }x=3.\;\text{ If }p(1)=6\text{ and }p(3)=2\text{, then }p'(0)= \_\_

    The desired polynomial is a cubic: . p(x) \:=\:ax^3 + bx^2 + cx + d
    Its derivative is: . p'(x) \:=\:3ax^2 + 2bx + c


    We are given:

    . . \begin{array}{ccccc}p(1) = 6\!:& a + b + c + d &=& 6 \\ \\[-4mm] p(3) = 2\!:& 27a + 9b + 3c + d &=&2 \\ \\[-4mm] p'(1) = 0\!: & 3a + 2b + c &=&0 \\ \\[-4mm] p'(3) = 0\!: & 27a + 6b + c &=&0 \end{array}


    Solve the system of equations: . a = 1,\;b = -6,\;c = 9,\;d = 2


    The cubic is: . p(x) \:=\:x^3 - 6x^2 + 9x + 2
    . . Its derivative is: . p'(x) \:=\:3x^2 - 12x + 9

    Therefore: . p'(0) \:=\:3(0^2) - 12(0) + 9 \:=\:9

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