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Thread: Relative Maximums and Minimums

  1. #1
    Junior Member Godfather's Avatar
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    Relative Maximums and Minimums

    let $\displaystyle P(x)= x^4+ax^3+bx^2+cx+d$
    The graph of y=P(x) is symmetric with respect to the Y Axis, the relative maximum at (0,1) and has an absolute minimum at (q,-3).

    (A). Determine the values of a,b,c and d and using these values write an expression for P(x)
    (B). find all possible values of q
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  2. #2
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    skeeter's Avatar
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    Quote Originally Posted by Godfather View Post
    let $\displaystyle P(x)= x^4+ax^3+bx^2+cx+d$
    The graph of y=P(x) is symmetric with respect to the Y Axis, the relative maximum at (0,1) and has an absolute minimum at (q,-3).

    (A). Determine the values of a,b,c and d and using these values write an expression for P(x)
    (B). find all possible values of q
    P(x) is symmetrical to the y-axis ... that should tell you what the coefficients "a" and "c" are.

    max at (0,1) tells you two things ... the value of "d" and that P'(x) = 0

    abs min at (q, -3) is also at (-q, -3) ... graph is symmetrical.
    also, P'(x) = 0 at x = q and x = -q.

    work on it.
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  3. #3
    Senior Member vincisonfire's Avatar
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    $\displaystyle
    P(x)= x^4+ax^3+bx^2+cx+d
    $
    Extremum
    $\displaystyle
    \frac{dP}{dx}= 4x^3+3ax^2+2bx+c
    $
    Relative maximum at (0,1) and has an absolute minimum at (q,-3).
    Since the function is symmetric there is also an absolute minimum at (-q,-3).
    $\displaystyle
    \frac{dP}{dx}= \alpha x (x-q)(x+q) = \alpha (x^3 - q^2 x)
    $
    Deduce that $\displaystyle \alpha = 4 , a = c = 0 \text { and } 2b = -\alpha q^2 = -4q^2 \implies b = -2q^2 $
    $\displaystyle
    P(x)= x^4-2q^2 x^2+d
    $
    $\displaystyle
    P(0)= 0^4-2q^2 0^2+d = d = 1
    $
    $\displaystyle
    P(q)= q^4-2q^4+1 = -3
    $
    $\displaystyle
    P(q)= -q^4 = -4
    $
    $\displaystyle
    P(q)= q^2 = 2
    $
    -2 doesn't work here that why we omit the sign.
    $\displaystyle
    P(q)= q = \pm \sqrt{2}
    $
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  4. #4
    Junior Member Godfather's Avatar
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    how did u deduce that formula after the first derivitive? where did the alpha come in? how did u do that?
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  5. #5
    Senior Member vincisonfire's Avatar
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    $\displaystyle
    \frac{dP}{dx}= \alpha x (x-q)(x+q) = \alpha (x^3 - q^2 x)
    $
    When x is equal to $\displaystyle q $ or $\displaystyle -q $, it is given that the first derivative is equal to 0 since there are local minima there.
    Therefore, the polynomial will have the form $\displaystyle \frac{dP}{dx}= f(x)(x-q)(x+q) $ so that it equals 0 when x is equal to $\displaystyle q $ or $\displaystyle -q $.
    Then you get that $\displaystyle \frac{dP}{dx}= g(x)(x)(x-q)(x+q) $ since there is a local max at x=0.
    You know that your polynomial is order three. Therefore, g(x) = constant.
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