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

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

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Answer 1 · 2017-10-10T05:15:29

The solution is $w \in (- \infty, - 2) \cup (0, 2)$ $w in (-oo,-2) uu (0,2)$

Explanation:

Let's factorise the inequality

$2 w^{3} - 8 w < 0$ $2w^3-8w<0$

$2 w (w^{2} - 4) < 0$ $2w(w^2-4)<0$

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

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

Let's build the sign chart

$a a a a$ $color(white)(aaaa)$ $w$ $w$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 2$ $-2$ $a a a a$ $color(white)(aaaa)$ $0$ $0$ $a a a a a a$ $color(white)(aaaaaa)$ $2$ $2$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $w + 2$ $w+2$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $w$ $w$ $a a a a a a a a a$ $color(white)(aaaaaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $w - 2$ $w-2$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (w)$ $f(w)$ $a a a a a a a$ $color(white)(aaaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

Therefore,

$f (w) < 0$ $f(w)<0$ , when $w \in (- \infty, - 2) \cup (0, 2)$ $w in (-oo,-2) uu (0,2)$

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

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

1 Answer

Explanation:

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