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Answer 1 · 2018-01-12T14:23:49

#x in[-1;3]#

#∣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+2x+1>=0 and |x|<=1#

#(x+1)^2>=0 and |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]#