which function has a removable discontinuity x-2/x^2-x-2, x^2-x+2/x+1, 5x/1-x^2, 2x-1/x

Question

which function has a removable discontinuity

x-2/x^2-x-2,

x^2-x+2/x+1,

5x/1-x^2,

2x-1/x

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Madeline 4 days 2021-09-15T00:02:35+00:00 1 Answer 0

Answers ( )

    0
    2021-09-15T00:03:56+00:00

    Answer:

    \frac{x-2}{x^2-x-2}

    Step-by-step explanation:

    A removable discontinuity is when there is a hole in your graph. This is usually because one X value has been canceled out. Most of the time, it takes factoring to figure out if there is a removable discontinuity when looking at an equation.

    First, look at the numerator x-2 . This can’t be factored any further. However, x^2-x-2 can be factored since it is a trinomial (has three terms) .

    For the purposes of this example, you may want to think about it as

    1x^2 -1x-2

    To factor, multiply the the outside coefficients

    1 x -2 = -2

    Now take the middle coefficient (-1) and ask yourself what two numbers multiply to make -2, but still add to be -1.

    -2 x 1 = -2

    -2 + 1 = -1

    So in factored form, the equation is

    \frac{x-2}{(x-2)(x+1)}

    Since you have x-2 on both top and bottom, that can be canceled out. x – 2 would be your removable discontinuity in this situation.

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