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avatar+1123 

Find the domain of the function
\[g(x) = \frac{x^3 + 11x - 2}{|x - 3| - |x + 1|}.\]

 Aug 9, 2023
 #1
avatar+126971 
+1

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)

 

 

cool cool cool

 Aug 9, 2023
 #2
avatar+36 
0

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, ∞)

 Aug 9, 2023

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