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March 28, 2017

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Question – 1: The gradient of the graph of f(x) at any point is,ax^2-12x + 9, and the point,(2,-3),is an inflection point of f(x).

(a) Find the value of the real constant, a. Find the function f(x).

(b) Find the stationary points of the function, then specify the intervals on which the function is increasing and those on which the function is decreasing

(c) Find the intervals on which f(x) is concave up, and those on which it is concave down.

  • Maths - ,

    1. Integrate the gradient function
    ax^2-12x + 9 to get f(x), except for an arbitary constant C.

    f(x) = (ax^3)/3 -6x^2 +9x + C

    If (2,-3) is an inflection point, the second derivative of f(x) must be zero there.

    2ax -12 = 0 @ x=2
    ax = 6
    a = 3

    Use f(x) and the statement that y=-3 at x=2 to solve for C

    (b) Stationary points occur where the derivative of f(x) is zero.

    (c) f(x) is concave down where f''(x) < 0 and concave up where f''(x) >0

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