Divide the problem into pieces
When x < 0, 2^x - 1 is always negative and 4 - 2^x always positive
From this, the equation becomes:
-1(2^x-1) + 4-2^x<=3 => -2*2^x <= -2 => 2^(x+1) >= 2 => x >= 0. Contradiction, so you can't have x < 0 .
At x = 0, you get (1-1) + (4-1) <=3, which is true => you *can* have x=0
When x is between 0 and 2, both bits are positive
2^x-1 + 4-2^x<=3 => The powers of 2 cancel out, leaving 3=3, which is true.
So x can be between 0 and 2.
At x = 2, you have
(4-1) + (0-0) <=3 This is true, so you can have x=2
When x >2, first bit is positive but second bit is negative
2^x-1 + -1(4-2^x) <=3 => 2^(x+1) <= 8 => x <= 2 Contradiction, so can't have x > 2 .
Therefore, the interval is [0, 2]
When x < 0, 2^x - 1 is always negative and 4 - 2^x always positive
From this, the equation becomes:
-1(2^x-1) + 4-2^x<=3 => -2*2^x <= -2 => 2^(x+1) >= 2 => x >= 0. Contradiction, so you can't have x < 0 .
At x = 0, you get (1-1) + (4-1) <=3, which is true => you *can* have x=0
When x is between 0 and 2, both bits are positive
2^x-1 + 4-2^x<=3 => The powers of 2 cancel out, leaving 3=3, which is true.
So x can be between 0 and 2.
At x = 2, you have
(4-1) + (0-0) <=3 This is true, so you can have x=2
When x >2, first bit is positive but second bit is negative
2^x-1 + -1(4-2^x) <=3 => 2^(x+1) <= 8 => x <= 2 Contradiction, so can't have x > 2 .
Therefore, the interval is [0, 2]