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    • Thread Starter

    I'm stuck on these two questions.

    1)f(x) ≡ x^2 + 16/x, x ≠ 0.

    a) Find f ′(x).

    b) Find the coordinates of the stationary point of the curve y = f(x) and determine
    its nature.



    f(x) = 4x^3 + ax^2 − 12x + b.
    Given that a and b are constants and that when f(x) is divided by (x + 1) there is a
    remainder of 15,
    a find the value of (a + b).
    Given also that when f(x) is divided by (x − 2) there is a remainder of 42,
    b find the values of a and b,
    c find the coordinates of the stationary points of the curve y = f(x).

    (a point to anyone who doesnt know what " x^2 or x^3 " etc is - it means x to the power of 2 , 3 i,e x squared, cubed etc :P

    a) Remember (16/x)= 16x^-1

    b) Stationary points occur when f ' (x) = 0. Find the value of x at this point and the corresponding value of y.
    'Nature' refers to whether the stationary point is a min. max. or point of inflexion.
    Find f ''(x) and insert the value for x:
    If f '' (x) is:
    negative - max. point
    positive - min. point
    zero - probably a point of inflexion
    • TSR Group Staff

    TSR Group Staff
    "Given that a and b are constants and that when f(x) is divided by (x + 1) there is a remainder of 15"

    This tells you that f(-1) = 15. You can do the division to find that out too.
    f(x) = 4x3 + ax2 - 12x + b
    15 = -4 + a + 12 + b
    7 = a + b

    If you apply the same logic with the next set of info (ie. f(2) = 42), you should end up with a pair of very simple simultaneous equations. Once you have a and b worked out, you can state f(x) in terms of x only. Then you can work out f'(x) = 0 for the stationary points.
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Updated: February 1, 2006

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