Mathematics, 10.03.2020 17:23 drainy0uandthefish

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

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Answer from: antcobra

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.

Answer from: Quest

you are saying it as if a fact, not a question, but yes, you would be correct? ; )

step-by-step explanation:

Answer from: Quest

the range of a function is the range of y-values for which it is defined, or exists. given that we are only looking at the y-values, we can already discount the first two answers as they deal with the domain of the function (range of x-values).

now, what we need to do is find the smallest and highest y-values for which the function is defined. if you'd like, you could take a ruler and start at the very bottom of the graph, find the lowest point, then move the ruler higher and higher until you reach the highest point. doing so either in your head or with an actual ruler, we can see that the lowest y-value is -5 and the highest y-value is -1.

the question now remains as to whether they are included in the range (fourth answer) or excluded (third answer). the key here is to recognise whether they are marked with an open or a closed circle. since both (-2, -5) and (2, -1) are marked with a closed circle, they are points that exist on the graph and are thus included in the range (note that if they were marked with an open circle like (4, -2) then that would signify that there is a break in the graph at that point).

thus, the fourth answer, (y | -5 ≤ y ≤ -1), is correct.

also note that usually it is good practice to scan your eye inbetween the highest and lowest point for any breaks in the graph - we do have one here at (4, -2), however -2 is still included in the range because there is another point that is included in the graph with the same y-value, ie. point (1, -2).

i hope that but if you require any further explanation don't hesitate to comment below.

Me if you can and if you finished this lesson on edg- comment you.

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