find domain and range
4x²y² - 3x² + y + 2 = 0
x²(4y² - 3) + y + 2 = 0
x² = -(y + 2) / (4y² - 3)
x² = (y + 2) / (3 – 4y²)
The lhs, x², must always be positive, therefore, so also must be the rhs.
The denominator on the rhs is (3 – 4y²), which is positive for |y| < √(3)/2.
The numerator on the rhs, (y + 2), is also positive for |y| < √(3)/2, and provided y ≥ -2.
The denominator on the rhs is (3 – 4y²), which is negative for |y| > √(3)/2.
The numerator on the rhs, (y + 2), is also negative for |y| > √(3)/2, provided y ≤ -2.
Writing Q = N/D, where N = (y + 2) and D = (3 – 4y²), i.e. Q = (y + 2) / (3 – 4y²).
+ – + –
Number line: ----|-------------------|-----------|----------|--------------------------
-2 -1.7321 0 1.7321
|
y
|
N
|
D
|
Q
|
Sign
|
|
-3
|
-1
|
-33
|
1/33
|
+
|
|
-1.8
|
0.2
|
-9.96
|
-2/99.6
|
–
|
|
0
|
2
|
3
|
2/3
|
+
|
|
1.8
|
3.8
|
-9.96
|
-38/99.6
|
–
|
The range on y then is: (-ꝏ, -2) U (-√(3)/2, √(3)/2).
We can rearrange the expression as a quadratic in y, i.e.
4x²y² + y + (2 – 3x²) = 0, or,
ay² + by + c = 0, where
a = 4x², b = 1, c = 2 – 3x²
The y-quadratic has real roots for a positive value for its discriminant. i.e. we need b^2 – 4ac ≥ 0.
1 – 4(4x²)(2 – 3x²) ≥ 0
1 – 16x²(2 – 3x²) ≥ 0
1 – 32x² + 48x44 ≥ 0
The solution to this quartic equation is,
x = ±0.1813, x = ±0.7961
We can rewrite the quartic equation as,
Q = (x – 0.1813)(x + 0.1813)(x – 0,7961)(x + 0.7961)
+ – + – +
Number line: ----|-------------------------------|---------|--------|-----------------------------------------|--------
-0.7961 -0.1813 0 0.1813 0.7961
|
x
|
Q
|
Sign
|
|
-0.8
|
0.00966
|
+
|
|
-0.5
|
-0.0814
|
–
|
|
0
|
0.02022
|
+
|
|
0.5
|
-0.0814
|
–
|
|
0.8
|
0.00966
|
+
|
We see that y has real values for a positive discriminant, which happens when the expression Q, in the Table, is greater than or equal to zero, as shown by the +-signs in the table.
Thus the domain of x is given by: (-ꝏ, -0.7961) U (-0.1813, 0,1813) U (0.7961, ꝏ)