How do you solve #28x ^ { 3} + ( 1- 3x ^ { 3} ) \geq ( x - 2) ^ { 3} + 35x ^ { 2}#?

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Answer 1 · 2018-01-17T20:42:12

#-0.612<=x<=0.446, x>=1.374#. (3 d.p)

simplify equation so that #0# is at RHS:

#28x^3 + (1-3x^3) = 28x^3 + 1 -3x^3#

#=25x^3 + 1#

#(x-2)^3 = x^3 + (3*-2*x^2) + (3*4*x) + (-2)^3#

#= x^3 -6x^2 + 12x - 8#

#(x-2)^3+35x^2 = x^3-6x^2+35x^2+12x-8#

#= x^3 + 29x^2 + 12x-8#

#28x^3+(1-3x^3) >= (x-2)^3 + 35x^2#

#25x^3+1>=x^3+29x^2+12x-8#

#24x^3+1 >=29x^2+12x-8#

#24x^3 >= 29x^2+12x-9#

#24x^3 - (29x^2+12x-9) >=0#
#24x^3 - 29x^2 - 12x + 9 >=0#

then input into a graph:

desmos.com/calculator

the roots (#x-#intercepts) are where #24x^3 - 29x^2 - 12x + 9 =0#

the values where the graph go above #0# are between where #x=-0.612# and #x = 0.446#, and to the right of where #x = 1.374#

in inequality notation, #-0.612<=x<=0.446, x>=1.374#.

(all figures to #3# decimal places)