The curve is a cubic with a maximum and minimum point. Graphically, two tangential lines can be drawn to touch the extrema. The vertical space between the two parallel lines represents the range of k.
f'(x)=3x^2-2x-1=(3x+1)(x-1)=0 at the extrema: x=-1/3 and 1 are the x coords and f(1)=-15/4; f(-1/3)=-277/108.
When f(x)=k, x^3-x^2-x-11/4=k so x^3-x^2-x=k+11/4. Clearly one answer is k=-11/4 because we get:
x(x^2-x-1)=0 and the three solutions are x=0, (1±√5)/2. But this is only one solution.
The range of k is between f(1) and f(-1/3), but does not include these values.
The picture shows the curve trapped between the parallel lines. These lines are at f(-1/3)=-277/108 and f(1)=-15/4. So we can write -277/108>k>-15/4. (The range is 32/27 units). Between these limits there are three solutions because a horizontal line cuts the curve in three places. Where the curve intersects the vertical axis is the point identified earlier where x=0 and ½(1±√5) and k=-2.75 (-11/4). You can also see that the integer value k=-3 is within the range.