Calculus

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Find a cubic function f(x) = ax^3 + bx^2 + cx + d that has a local maximum value of 3 at x = −2 and a local minimum value of 0 at x = 1.

  • Calculus -

    At the local min/max points, the derivative is zero.
    f' = 3ax^2 + 2bx + c = 0
    Plug in x = -3 and x = 2, and that gives you two equations.

    Also, use f = ax^3 + bx^2 + cx + d
    Plug in the values (x = -3, f = 3) and (x = 2, f = 0).
    That gives you two more equations.

    Now solve the four simultaneous linear equations in a, b, c, and d.

  • Calculus -

    f ' (x) = 3ax^2 + 2bx + c
    f '(-2) = 0
    12a -4b + c = 0 (#!1
    f ' (1) = 0
    3a + 2b + c = 0 (#2)

    #1 - #2 ----> 9a - 6b = 0
    or 3a = 2b

    also (-2,3) lies on original
    -8a + 4b - 2c + d = 3
    and (1,0) lies on it
    a + b + c + d = 0
    subtract those two equations
    9a -3b+3c = -3 ,or
    3a - b + c = -1 , (#3)

    #2 - #3 :
    3b = 1
    b = 1/3 , but a = 2b/3 = (2/3)(1/3) = 2/9

    in #3
    3(2/9) -1/3 + c = -1
    c = -4/3

    in a+b+c+d = 0
    2/9 + 1/3 - 4/3 + d = 0
    d = 7/9

    so f(x) = (2/9)x^3 + (1/3)x^2 - (4/3)x + 7/9

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