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a) For the cubic to have no stationary points the derivative cannot equal 0 (dy/dx =\= 0).This means that the discriminant must equal less than 0. For this to occur, assuming ax^3+bx^2+cx+d, both a and c must be either + or - but they cannot be different. B must also be considerably lower than a and c as b^2 must be less than 4ac. An example would be 4x^3-2x^2+3x.
b) For a cubic to have stationary points at x=2 and 5 the derivative must equal (x-2)(x-5). This can then be expanded and anti-differentiated to become x^3/3-3.5x^2+10x+d however d can be any number.
c) In this case the derivative must equal (x+1)(x+d). This is because there will be a stationary point at x= -1 and another at d. To make x=-1 the local minimum d must be more than 1 so that the other turning point is a maximum. An example would be x^3/3+2x^2+3x.
d) The derivative must equal (x+2)(x-4) to give stationary points at x=-2 and 4. When expanded and anti-differentiated this becomes x^3/3-x^2-8x. Though this gives a local minimum at x=4 and maximum at x=-2 so the graph must be flipped over the x axis and so the final solution becomes -(x^3/3-2x^2-8x).