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Math Help - Relative Maximums and Minimums

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

    let 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 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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    <br />
P(x)= x^4+ax^3+bx^2+cx+d<br />
    Extremum
    <br />
\frac{dP}{dx}= 4x^3+3ax^2+2bx+c <br />
    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).
    <br />
\frac{dP}{dx}= \alpha x (x-q)(x+q) = \alpha (x^3 - q^2 x) <br />
    Deduce that \alpha = 4 , a = c = 0 \text { and } 2b = -\alpha q^2 = -4q^2 \implies b = -2q^2
    <br />
P(x)= x^4-2q^2 x^2+d<br />
    <br />
P(0)= 0^4-2q^2 0^2+d = d = 1<br />
    <br />
P(q)= q^4-2q^4+1 = -3<br />
    <br />
P(q)= -q^4 = -4<br />
    <br />
P(q)= q^2 = 2<br />
    -2 doesn't work here that why we omit the sign.
    <br />
P(q)= q = \pm \sqrt{2}<br />
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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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    <br />
\frac{dP}{dx}= \alpha x (x-q)(x+q) = \alpha (x^3 - q^2 x)<br />
    When x is equal to  q or  -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  \frac{dP}{dx}= f(x)(x-q)(x+q) so that it equals 0 when x is equal to  q or  -q .
    Then you get that  \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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