# Question 84c3c

Jan 12, 2018

x in[-1;3]

#### Explanation:

∣x^2−1∣≤2x+2 

We must divide the equation in two different parts for this we must discover the zeros of x^2−1:

It is easy to see that x^2−1=(x+1)(x-1), so the zeros are 1 and -1.

Between the zeros the expression is negative so its modulus is:
$- {x}^{2} + 1.$

Solving:
-x^2+1≤2x+2 and |x|<=1

-x^2-2x-1≤0 and |x|<=1

${x}^{2} + 2 x + 1 \ge 0 \mathmr{and} | x | \le 1$

${\left(x + 1\right)}^{2} \ge 0 \mathmr{and} | x | \le 1$

${\left(x + 1\right)}^{2} \ge 0 \forall x \in \mathbb{R} \mathmr{and} | x | \le 1$

x in [-1; 1]

Outside the zeros:
x^2−1≤2x+2 and |x|>1

x^2-2x−3≤0 and |x|>1

Temos de calcular os zeros desta expressão:

$x = \frac{2 \pm \sqrt{4 - 4 \times 1 \times \left(- 3\right)}}{2}$

$x = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2} = 1 \pm 2$

$x = - 1 \mathmr{and} x = 3$

so:

x in ]-1; 3] and x notin ]-1; 1[

This is equivalent to

]1; 3]

Joining both cases we obtain

x in[-1;3]#