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Given: $f(x) =x^3 - 3x^2 + 4$

  1. Calculate $f(-1)$, and hence solve the equation $f(x)=0$
  2. Determine $f'(x)$
  3. Sketch the graph of $f$ neatly and clearly, showing the co-ordinates of the turning points as well as the intercepts on both axes.
  4. Determine the co-ordinates of the points on the graph of $f$  where the gradient is $9$.
in Grade 12 Maths by Bronze Status (5.9k points) | 15 views

1 Answer

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1. f(-1) = 0

x+1 is therefore a factor to divide into cubic using long division, but put 0x as a place holder(x^3 - 3x^2 + 0x +4)

= (x+1)(x^2 -4x +4)= 0

  (x+1)(x-2)^2 = 0

    x= -1 or 2 twice

2. dy/dx = 3x^2 -6x

3. 3x^2 -6x=0

    x(x-2)=0

    x=0 or 2

subst into cubic, y=4 or 0, giving (0;4) and (2;0)

4. 3x^2-6x=9

   (x+1)(x-3)=0

      x= -1 or 3

subst into cubic, y=0 or 4, giving (-1;0) and (3;4)
by Diamond (82.9k points)

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