-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