# Prove that if $-2 \leq x_0 \leq 2$, then $-2 \leq 3x_0 - x_0^3 \leq 2$.

Given $$-2 \leq x_0 \leq 2$$, I have to prove $$-2 \leq x_n \leq 2$$, where $$x_n = 3x_{n - 1} - x_{n - 1}^3$$ and $$n \in \mathbb{Z}_+$$. I thought this problem can be solve by mathematical induction, but I can't prove the initial step, i.e., $$n = 1$$.

I try this two ways:

1st. If $$-2 \leq x_0 \leq 2$$, then $$-6 \leq 3x_0 \leq 6$$ and $$-8 \leq x_0^3 \leq 8$$. But, this provied the inequality

$$$$-14 \leq 3x_0 - x_0^3 \leq 14,$$$$

that its true.

2nd. I try prove it separately, with $$0 \leq x_0 \leq 2$$ first, but its wrong too.

How can I prove it?

• Can you fix the title. The title reads that $$0\le3x-x^3$$ it should read $$-2\le3x-x^3$$ Nov 3 '20 at 2:38
• Yes, that its the problem Nov 3 '20 at 3:13
• how did you arrive at the interval that you need to use for your inductive step? Nov 3 '20 at 3:16
• Its given by the problem. Nov 3 '20 at 3:17
• @MathKeepsMeBusy $3(-2)-(-2)^3=-6+8=2$ Nov 3 '20 at 5:21

$$f(x)=3x-x^3 \implies f'(x)=3-3x^2, f'(x)=0 \implies x=0,\pm 1$$ So $$f_{max}=f(1)=2, f(2)=-2, f(0)=0 \implies -2\le f(x) \le 2$$

There is an elementary way to prove this using AM-GM. If $$x^2<3$$, then

$$\left( x(3-x^2) \right)^2 =\frac{1}{2} 2x^2 \cdot (3-x^2) \cdot (3-x^2)\\ \leqslant \frac{1}{2} \left(\frac{2x^2+3-x^2+3-x^2}{3} \right)^3 =4.$$

If $$x^2\ge 3$$, then $$\left( x(3-x^2) \right)^2 =x^2 \cdot (x^2-3)^2 \leqslant 4\cdot 1^2=4.$$

• How can I prove the inferior limit in this ways? Nov 3 '20 at 3:14
• I edited. Just square everything. Nov 3 '20 at 3:20
• Your forgot square the last 2. Nov 3 '20 at 3:32
• Doesn't AM-GM require all terms be positive? What if $3-x^2 < 0$? Nov 3 '20 at 4:08
• Thanks, tajiri_numero_1 @fleablood Good point, I just fixed. Thank you. Nov 3 '20 at 4:20

Prove that $$f(x)+2$$ is positive, and $$f(x)-2$$ is negative. Hint: factors