Question

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

Answer 1

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.