Briefly, if f(x)=k, x^3-x^2-x=k+11/4, so if k=-11/4, we have 3 solutions x(x^2-x-1)=0, x=0, (1±√5)/2. Using calculus you can discover where the max and min of f(x) are and you can use these to work out the range of k. Graphically, the cubic has two "humps" and the range of k is determined by the vertical distance between them, because if you slide the x axis between the extrema it cuts the curve in 3 places, which are the solutions. Post the question fully and I'll look out for it. In the meantime I hope this helps.
