How do you solve and write the following in interval notation: #x^3 + 3x<4x^2 #?

1 Answer
Binayaka C.
Apr 12, 2017

Solution : # x <0 and 1 < x < 3 #. In interval notation: # ( -oo ,0) uu (1 ,3) #

Explanation:

#x^3 + 3x <4x^2 or x^3 + 3x -4x^2 < 0 or x(x^2 -4x +3) < 0 # or

#x (x-1)(x-3) < 0#. Critical points are # x=0 , x=1 , x=3#

When #x < 0 # ::: #x (x-1)(x-3)# is negative i.e #<0#

When # 0 < x < 1 # ::: #x (x-1)(x-3)# is positive i.e # > 0#

When # 1 < x < 3 # ::: #x (x-1)(x-3)# is negative i.e # < 0#

When #x > 3 # ::: #x (x-1)(x-3)# is positive i.e #>0#

So Solution : # x <0 and 1 < x < 3 # in interval notation # ( -oo ,0) uu (1 ,3) # graph{x^3-4x^2+3x [-10, 10, -5, 5]}. The graph also confirms above result #[Ans]

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