Let [ ] denote the greatest integer function. If the domain of the function f(x) = ⁻¹ ( 4x+2[x] 3 ) is [ , ] , then 12( + ) is equal to:

Step-by-step solution

For the domain of f(x) = ⁻¹ ( 4x+2[x] 3 ) , the argument must satisfy: -1 4x+2[x] 3 1 -3 4x + 2[x] 3 Let x = I + f , where I = [x] Z and f = x [0, 1) . Substituting x = I + f into the inequality: -3 4(I + f) + 2I 3 -3 6I + 4f 3 Since 0 f If I 1 , 6I + 4f 6(1) + 0 = 6 > 3 , which gives no solution. If I -2 , 6I + 4f Thus, the only possible values for I are 0 and -1 . Case 1: I = 0 -3 6(0) + 4f 3 -3 4f 3 - 3 4 f 3 4 Since f [0, 1) , we get 0 f 3 4 . Therefore, x = I + f [0, 3 4 ] . Case 2: I = -1 -3 6(-1) + 4f 3 3 4f