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-2 < x²-4x+1 < 1
This is the same as
-2 < x²-4x+1 AND x²-4x+1 < 1
We find the critical numbers for each
First we find the critical numbers for
-2 < x²-4x+1 which is the same as
x²-4x+1 > -2
x²-4x+3 > 0
(x-1)(x-3) > 0
Critical numbers are 1 and 3
Next we find the critical numbers for
x²-4x+1 < 1
x²-4x < 0
x(x-4) < 0
Critical numbers are 0 and 4
We place all four critical numbers on a number line:
----------o--o-----o--o---------
-3 -2 -1 0 1 2 3 4 5 6 7
We test the interval x < 0 with test value -1
-2 < x²-4x+1 < 1
-2 < (-1)²-4(-1)+1 < 1
-2 < 1+4+1 < 1
-2 < 6 < 1
False so we don't include x < 0
----------o--o-----o--o---------
-3 -2 -1 0 1 2 3 4 5 6 7
We test the interval 0 < x < 1 with test value 0.5
-2 < x²-4x+1 < 1
-2 < (0.5)²-4(0.5)+1 < 1
-2 < 0.25-2+1 < 1
-2 < -0.75 < 1
That's true so we include 0 < x < 1
----------o==o-----o--o---------
-3 -2 -1 0 1 2 3 4 5 6 7
We test the interval 1 < x < 3 with test value 2
-2 < x²-4x+1 < 1
-2 < (2)²-4(2)+1 < 1
-2 < 4-8+1 < 1
-2 < -3 < 1
That's false so we do not include 1 < x < 3
----------o==o-----o--o---------
-3 -2 -1 0 1 2 3 4 5 6 7
We test the interval 3 < x < 4 with test value 3.5
-2 < x²-4x+1 < 1
-2 < (3.5)²-4(3.5)+1 < 1
-2 < 12.25-14+1 < 1
-2 < -0.75 < 1
That's true so we include 3 < x < 4
----------o==o-----o==o---------
-3 -2 -1 0 1 2 3 4 5 6 7
We test the interval x > 4 with test value 5
-2 < x²-4x+1 < 1
-2 < (5)²-4(5)+1 < 1
-2 < 25-20+1 < 1
-2 < 6 < 1
That's false so we do not include x > 4
-------------------
Since the inequality is STRICT inequality, the critical numbers
cannot be solutions, and the graph of the solution set is
----------o==o-----o==o---------
-3 -2 -1 0 1 2 3 4 5 6 7
which in interval notation is:
(0,1) U (3,4)
Edwin
