How do you solve #2w^3-8w<0# using a sign chart?

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Answer 1 · 2017-10-10T05:15:29

The solution is #w in (-oo,-2) uu (0,2)#

Let's factorise the inequality

#2w^3-8w<0#

#2w(w^2-4)<0#

#2w(w+2)(w-2)<0#

Let #f(w)=2w(w+2)(w-2)#

Let's build the sign chart

#color(white)(aaaa)##w##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaa)##0##color(white)(aaaaaa)##2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##w+2##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##w##color(white)(aaaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##w-2##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(w)##color(white)(aaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(w)<0#, when #w in (-oo,-2) uu (0,2)#

graph{2x^3-8x [-14.24, 14.24, -7.12, 7.12]}