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Solve the inequality x^3 >= 4x.

 Nov 26, 2020
 #1
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x^3 ≥  4x

 

x^3 - 4x ≥  0     (1)

 

x  ( x^2  - 4)  ≥  0

 

x  ( x + 2)  ( x - 2) ≥  0

 

Setting each factor  =  0   and solving for x  produces  three test points

 

x =  0     x = -2     x   = 2

 

We have the  intervals

 

(-inf, 2]       [ -2, 0 ]      [ 0, 2]      [ 2, inf  )

 

Picking a point in each interval   and testing this in (1)

 

x = -3    (-3)^3 - 4(-3)   =   -27 + 12  =   -15    and this is not ≥  0....so the first interval doesn't work

x = -1   (-1)^3  - 4(-1)  =  -1 + 4  = 3    and this is ≥ 0....so the second interval works

x = 1    (1)^3  - 4(1) =   -3    and this is not ≥ 0....so  this interval does not work

x = 3    (3)^3  - 4(3)  =  15     and this is ≥  0  ...so this interval works

 

The solution   is   x =  [-2, 0 ] U  [ 2, inf )

 

 

cool cool cool

 Nov 26, 2020

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