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diamong [38]
4 years ago
6

Simplify the following rational expression. List all of the excluded values, and classify each as a removable discontinuity, or

a nonremovable discontinuity. please help
-2/x^2-4 +x-1/x^2-2x
Mathematics
1 answer:
Nana76 [90]4 years ago
4 0

Answer:

1.           \dfrac{x+1}{x(x+2)}

2. The excluded values are x = 0 (not removable), x = -2 (not removable), and x = 2 (removable)

Explanation:

The excluded values are those that make any denominator 0.

1. Excluded values for      

                                    -\dfrac{2}{x^2-4}

     x^2-4=0\implies x^2=4\implies x=\pm2

2. Excluded values for

                                    \dfrac{x-1}{x^2-2x}

      x^2-2x=0\implies x(x-2)\implies x=0\text{ }and\text{ } x=2

3. Join both sets

  • The excluded values are x = 0, x = -2, and x = 2

4. Simplify

         

             -\dfrac{2}{x^2-4}+\dfrac{x-1}{x^2-2x}=-\dfrac{2}{(x+2)(x-2)}+\dfrac{x-1}{x(x-2)}=\\ \\ \\ \dfrac{-2x+(x+2)(x-1)}{(x+2)(x-2)x}=\\ \\ \\ \dfrac{-2x+x^2+x-2}{(x+2)(x-2)x}=\dfrac{x^2-x-2}{(x+2)(x-2)x}=\dfrac{(x-2)(x+1)}{(x+2)(x-2)x}=\\ \\ \\ \dfrac{x+1}{(x+2)x}

  • At the end, the factor x - 2 was canceled, meaning that the excluded value x = 2 is removable.

  • x = -2 and x = 0 yet make the final expression undefined, thus they are not removable.
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