Question 1125739
your inequality is x^2 <= x^3


subtract x^2 from both sides of this equation to get:


0 <= x^3 - x^2.


find the 0 points of the graph by setting x^2 - x^2 = 0


factor out an x^2 to get x^2 * (x-1) = 0


solve for x to get x = 0 or x = 1.


those are the 0 points of the graph (y = 0).


now you want to look at the intervals where x < 0 and x > 1 and 0 < x < 1.


since we only care about the intervals where x >= 0, then the intervals we are interested in are 0 < x < 1 and x > 1.


since the graph is continuous, if it is above the x-axis, it will stay above until it crosses a 0 point and if it is below the x-axis, it will stay below until it crosses a 0 point.


pick a value in each interval to test.


when x = .5, x^2 = .25 and x^3 = .125, therefore x^2 greater than x^3.


when x = 2, x^2 = 4 and x^3 = 8, therefore x^2 less than x^3.


the solution is that x^2 <= x^3 in the interval of x >= 1.


the following graph confirms that.


<img src = "http://theo.x10hosting.com/2018/101004.jpg" alt="$$$" >