How do you solve $x^{5} + 9 x \geq 10 x^{3}$ $x^5+9x>=10x^3$ ?

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How do you solve $x^{5} + 9 x \geq 10 x^{3}$ $x^5+9x>=10x^3$ ?

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Answer 1 · 2016-12-15T12:35:27

The answer is $x \in [- 3, - 1] \cup [0, 1] \cup [3, + \infty [$ $x in [-3,-1] uu [0, 1] uu [3,+ oo [$

Explanation:

Let $f (x) = x^{5} - 10 x^{3} + 9 x$ $f(x)=x^5-10x^3+9x$

Let's rearrange the equation

$x^{5} - 10 x^{3} + 9 x \geq 0$ $x^5-10x^3+9x>=0$

$x (x^{4} - 10 x^{2} + 9) \geq 0$ $x(x^4-10x^2+9)>=0$

$x (x^{2} - 1) (x^{2} - 9) \geq 0$ $x(x^2-1)(x^2-9)>=0$

$x (x + 1) (x - 1) (x + 3) (x - 3) \geq 0$ $x(x+1)(x-1)(x+3)(x-3)>=0$

We can do a sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 3$ $-3$ $a a a a$ $color(white)(aaaa)$ $- 1$ $-1$ $a a a a$ $color(white)(aaaa)$ $0$ $0$ $a a a a$ $color(white)(aaaa)$ $1$ $1$ $a a a a$ $color(white)(aaaa)$ $3$ $3$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 3$ $x+3$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x + 1$ $x+1$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a a a a a a$ $color(white)(aaaaaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x - 1$ $x-1$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x - 3$ $x-3$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a a a$ $color(white)(aaaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $-$ $color(white)(aaa)$ $+$ $color(white)(aaa)$ $-$ $color(white)(aaa)$ $+$

So,

$f(x)>=0$ when $x in [-3,-1] uu [0, 1] uu [3,+ oo [$

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