+1348 Find the domain of the function
\[g(x) = \frac{x^3 + 11x - 2}{|x - 3| - |x + 1|}.\]
+129830 The function is not defined if the denominator = 0
Note that l x l = sqrt (x^2)
So
sqrt [ (x - 3)^2 ] - sqrt [ (x + 1)^2 ] = 0
sqrt [ (x - 3)^2 ] = sqrt [ ( x + 1)^2 ] square both sides
(x - 3)^2 = (x + 1)^2
x^2 - 6x + 9 = x^2 + 2x + 1
-6x + 9 = 2x + 1
9 - 1 = 6x + 2x
8 = 8x
x = 8/8 = 1
So x =1 makes the denominator = 0
So the domain is (-inf, 1) U ( 1, inf)
+36 We observe that this is a rational function in which the numerator is
a polynomial of degree 3 and this polynomial is defined for all real values.
So, this rational function will be undefined at the points where we will have,
denominator = 0
This gives us,
|x - 3| - |x + 1| = 0 ......... (1)
To remove the absolutes in equation (1), we will have following three intervals:
x ≤ -1, -1 ≤ x ≤ 3, x ≥ 3
If x ≤ -1, then using (1), we get
-(x - 3) - (-(x + 1)) = 0
- x + 3 + x + 1 = 0
4 = 0
This is not a true statement.
It means that we have no solution when x ≤ -1.
If -1 ≤ x ≤ 3, then using (1), we get
-(x - 3) - (x + 1) = 0
-x + 3 - x - 1 = 0
-2x + 2 = 0
2x = 2
x = 2/2
x = 1
It means that when -1 ≤ x ≤ 3, then the solution is given by
x = 1
If x ≥ 3, then using (1), we get
(x - 3) - (x + 1) = 0
x - 3 - x - 1 = 0
- 4 = 0
This is not a true statement.
It means that we have no solution when .
Therefore, the solution of the equation (1), is given by x = 1
Hence, the function g(x) is defined for all real values except x = 1.
Therefore, the domain of the function g(x) is (-∞, 1)U(1, ∞).
Domain (-∞, 1)U(1, ∞)
