How do you solve #x^3>=9x^2#?

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Answer 1 · 2016-12-27T00:55:43

#x in [9, oo)uu{0}#

#x^3>=9x^2#

#=> x^3-9x^2>=0#

#=> x^2(x-9)>=0#

#x^2(x-9)# has two roots: #x=0# and #x=9#. We will check what happens in the intervals on each side of them.

Case 1: #x in (-oo, 0)#

#=> x^2 > 0 and x-9 < 0#

#=> x^2(x-9) < 0#

Case 2: #x in (0, 9)#

#=> x^2>0 and x-9 < 0#

#=> x^2(x-9) < 0#

Case 3: #x in (9, oo)#

#=> x^2 > 0 and x-9 > 0#

#=> x^2(x-9) > 0#

Thus we have #x^2(x-9)>=0# if #x=0# or #x in [9, oo)#