|x^2 - 2| < 1 Please can anyone help? How to solve this absolute value problem?

↳Redirected from "Suppose that I don't have a formula for #g(x)# but I know that #g(1) = 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
2 Answers
Narad T.
Mar 15, 2018

The solution is #x in (-sqrt3,-1)uu(1, sqrt3)#

Explanation:

#|x^2-2|<1#

There are #2# solutions

#x^2-2<1# and #x^2-2> -1#

#x^2-3<0#,#=># #x in (-sqrt3, sqrt3)#

#x^2-1>0#, #=>#, #x in (-oo,-1)uu(1,+oo)#

You can prove theses solutions by a sign chart.

Combining the #2# solutions

#x in (-sqrt3,-1)uu(1, sqrt3)#

graph{|x^2-2|-1 [-6.24, 6.25, -3.117, 3.126]}

Cesareo R.
Mar 15, 2018

See below.

Explanation:

Squaring both sides

#(x^2-2)^2 < 1 rArr (x^2-2)^2 - 1 < 0 rArr (x^2-2-1)(x^2-2+1) < 0# or

#(x^2-3)(x^2-1) < 0# which is equivalent to

#{(x^2-3 < 0),(x^2-1>0):} rArr 1 < x^2 < 3#

the set

#{(x^2-3 > 0),(x^2-1<0):} rArr {x^2 < 1} nn {x^2 > 3}#

is an extraneous solution, is not feasible and was included due to the squaring operation.

Finally

#1 < x^2 < 3#

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