calculus

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1. Let f be the function that is defined for all real numbers x and that has the following properties.
(i) f''(x)=24x-18
(ii) f'(1)=-6
(iii) f'(2)=0
a. Find each x such that the line tangent to the graph of f at (x, f (x)) is horizontal
b. Write an expression for f (x)

  • calculus -

    f'(x) = 12x^2 - 18x + c
    for f'(1) = -6
    12 - 18 + c = -6
    c = 0

    so f'(x) = 12x^2 - 18x

    Then you say f'(2) = 0 , which does not verify with what I have so far.
    Did you mean f(2) = 0 ?
    I will assume you meant that.
    f(x) = 4x^3 - 9x^2 + k
    given f(2) = 0
    32 - 36 + k = 0
    k = 4

    f(x) = 4x^3 - 9x^2 + 4

    leaving up to you: .....

    a) .......
    tell me what you would do.

  • calculus -

    f''(x) = 24x - 18

    f'(x) = 12x^2 - 18x + C
    Since f'(1) = -6, then
    -6 = 12(1)^2 - 18(1) + C
    -6 = 12 - 18 + C
    -6 = -6 + C
    C = 0

    So f'(x) = 12x^2 - 18x

    f(x) = 4x^3 - 9x^2 + C
    Since f(2) = 0, then
    0 = 4(2)^3 - 9(2)^2 + C
    0 = 4(8) - 9(4) + C
    0 = 32 - 36 + C
    0 = -4 + C
    4 = C

    f(x) = 4x^3 - 9x^2 + 4

    For Part A, you know that f'(x) = 12x^2 - 18x
    f'(x) = 0
    12x^2 - 18x = 0
    6x(2x-3) = 0
    x = 0 or x = 3/2
    Just evaluate f(0) and f(3/2) to find the y-coordinates.

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