



 
Hi Katie, I am going to answer your question but for the expression x^{3}  2x^{2}. I think about this question in terms of the graph of the expression y = x^{3}  2x^{2}. I want to know when the graph lies above the xaxis (that is when x^{3}  2x^{2} > 0), when it is below the xaxis (that is when x^{3}  2x^{2} < 0) and when it intersects the xaxis (that is when x^{3}  2x^{2 }= 0). To determine when x^{3}  2x^{2 }= 0 I factor x^{3}  2x^{2} to obtain x^{3}  2x^{2} = x^{2}(x  2) and hence x^{3}  2x^{2} = x^{2}(x  2) = 0 when x = 0 and when x = 2. These two xvalues divide the xaxis into three pieces I now choose an xvalue in the leftmost piece, x < 0. I want an xvalue that is easy to calculate with so I choose x = 1. At x = 1, y = x^{3}  2x^{2} = (1)^{3}  2(1)^{2} = 1 2 = 3. Thus the graph of y = x^{3}  2x^{2} is below the xaxis at x = 1. The important point here is that the graph is thus below the xaxis for all x satisfying x < 0 since to go above the xaxis the graph would need to cross the xaxis and it doesn't touch the xaxis until x = 0. Now choose a point in the next piece of the axis, 0 < x < 2. Again I want an easy point so take x = 1. At x = 1, y = x^{3}  2x^{2} = 1^{3}  2(1)^{2} = 1 2 = 1. Thus the graph of y = x^{3}  2x^{2} is below the xaxis at x = 1 and thus below the xaxis for all x in 0 < x < 2 since it doesn't touch the xaxis again until x = 2. Finally choos a point in the remaining piece of the xaxis, 2 < x. This time I take x = 3 and at x = 3, y = x^{3}  2x^{2} = 3^{3}  2(3)^{2} = 27  18 = 9 and the graph is above the xaxis at x = 3 and therefore above the xaxis fo all x satisfying 2 < x. Thus my answer is x^{3}  2x^{2} ≥ 0 for 2 ≤ x and for x = 0. Just as a check here is the graph of y = x^{3}  2x^{2}. Penny  


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