Question #84c3c

1 Answer
Jan 12, 2018

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

Explanation:

∣x^2−1∣≤2x+2 x212x+2

We must divide the equation in two different parts for this we must discover the zeros of x^2−1x21:

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.x2+1.

Solving:
-x^2+1≤2x+2 and |x|<=1x2+12x+2and|x|1

-x^2-2x-1≤0x22x10 and |x|<=1#

x^2+2x+1>=0 and |x|<=1x2+2x+10and|x|1

(x+1)^2>=0 and |x|<=1(x+1)20and|x|1

(x+1)^2>=0 AA x in RR and |x|<=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=(2+-sqrt(4-4xx1xx(-3)))/2

x=(2+-sqrt(4+12))/2=(2+-sqrt(16))/2=(2+-4)/2=1+-2

x=-1 or 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]