Inequalities 2

Domain of given inequality-

$\begin{aligned}8-x^{3}&≥0\\ x^{3}&≤8\\ x&≤2\\ D&\in(-\infty,2)\end{aligned}$

Since $\sqrt{8-x^{3}}$ is always positive for the given domain,the original expression can be re-written as,

$|x-2|-|x|≥0$

The given expression can be solved by dividing it into different cases:

CASE 1

$\begin{aligned}X_1≤-2,\\ |x+2|&=-(x+2)\\ |x|&=-x \\ \\ -x-2+x&≥0 \\-2&≥0 \\ \therefore x_1&\in\phi\end{aligned}$

CASE 2

$\begin{aligned}-2<X_2≤0,\\ |x+2|&=x+2\\ |x|&=-x \\ \\ x+2+x&≥0\\ 2x+2&≥0\\ x+1&≥0\\ x&≥-1\\x&\in[-1,\infty)\\x_2&\in[-1,\infty)\cap[-2,0] \\ x_2&\in[-1,0] \end{aligned}$

CASE 3

$\begin{aligned}X_3>0,\\ |x+2|&=x+2 \\ |x|&=x\\ \\ x+2-x&≥0 \\ 2&≥0 \\ \therefore x_3&\in(0,\infty)\end{aligned}$

$\begin{aligned}X&\in x_1\cup x_2\cup x_3\\ X&\in \phi \cup [-1,0] \cup (0,\infty) \\ X&\in [-1,\infty)\\ \\ \therefore X_{final}&\in X \cap D\\ X_{final}&\in [-1,\infty)\cap (-\infty,2)\\ X_{final}&\in[-1,2)\end{aligned}$

$\therefore a=-1,b=2 \\|a|+b=1+2=\boxed{3}$

Let $f(x) = \dfrac {|x+2|-|x|}{\sqrt{8-x^3}}$ . We note that the denominator $\sqrt{8-x^3}\ge 0$ for $x \le 2$ and not real for $x>2$ . Therefore, $f(x)$ is defined for $x<2$ . Since $\sqrt{8-x^3}> 0$ for $x<2$ , $f(x) \ge 0$ , when its nominator $|x+2|-|x| \ge 0$ . There are three cases to be considered:

$|x+2|-|x| = \begin{cases} -x-2+x = - 2 \color{#D61F06} < 0 & & \text{for }x \in (-2, \infty) \\ x+2+x = 2x + 2 \ge 0 & \color{#3D99F6} \implies x \ge -1 & \text{for } x \in [2, 0) \\ x + 2-x = 2 \color{#3D99F6} > 0 & & \text{for } x \in [0, 2) \end{cases}$

$\implies \dfrac {|x+2|-|x|}{\sqrt{8-x^3}} \ge 0$ for $x \in [-1, 2)$ . $\implies |a|+b = |-1|+2 = \boxed{3}$