To solve f(x)>d, draw y=f(x) and y=d, identify their intersections, and select the portions above the line. For <, select below; include boundary points for <= or >=. The roots of f alone do not locate the intersections when d 0. For f(x)=x(x-1)(x+1), comparison with y=6 gives x3-x-6=(x-2)(x2+2x+3). The quadratic is (x+1)2+2>0, so there is one intersection at (2,6). The graph lies above y=6 exactly when x>2.

To solve \(\mathrm{f}(x)>d\), draw \(y=\mathrm{f}(x)\) and \(y=d\), identify their intersections, and select the portions above the line. For \(<\), select below; include boundary points for \(\le\) or \(\ge\). The roots of \(\mathrm{f}\) alone do not locate the intersections when \(d\ne0\).

For \(\mathrm{f}(x)=x(x-1)(x+1)\), comparison with \(y=6\) gives \(x^3-x-6=(x-2)(x^2+2x+3)\). The quadratic is \((x+1)^2+2>0\), so there is one intersection at \((2,6)\). The graph lies above \(y=6\) exactly when \(x>2\).

A horizontal comparison line

A horizontal comparison line