Question 1125739: for all real numbers x>0, we have x^2<= x^3
Found 2 solutions by ikleyn, Theo:
Answer by ikleyn(52818)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! 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.
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