Calculus

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Find the equation of a quartic polynomial whose graph is symmetric about the y-axis and has local maxima at (−1,1) and (1,1) and a y-intercept of -1.

  • Calculus -

    y = a x^4 + b x^3 + c x^2 + d x + e

    when x = 0, y = -1 so e = -1
    so
    y = a x^4 + b x^3 + c x^2 + d x - 1

    for max or min
    dy/dx = 0 = 4 a x^3 + 3 b x^2 + +2 c x + d

    when x = -1
    0 = 4 a (-1) +3b(1)+2c(-1)+d
    1 = a (1) + b(-1) + c(1) +d(-1) - 1

    when x = +1
    0 = 4a + 3 b + 2c + d
    1 = a+b+c+d-1
    --------------------------------
    start solving
    0 = -4a+3b-2c+d
    0 = +4a+3b+2c+d
    -----------------
    0 = 6b +2d

    2 = a-b+c-d
    2 = a+b+c+d
    -------------
    0 = -2b-2d

    0 = 6b +2d
    0 = -2b-2d
    -----------
    8 b = 0
    b=0
    then d = 0
    (in fact from symmetry I bet all odd terms are 0)
    now go back and get a and c

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