Find the dicontinuities of the function.
1 answer:
Answer:
Step-by-step explanation:
I'm assuming you meant to type in
because you can only have removable discontinuities where there is a rational (fraction) function. Begin by factoring both the numerator and denominator to
and cancelling out like terms would have us eliminating the (x + 3). That is where there is a removable discontinuity. It leaves a hole. The other discontinuity, (x + 1) doesn't cancel out so it is a non-removable discontuinity, which is a vertical asymptote.
The removable discontinuity is at -3. There is no y value at x = -3 (remember there's only a hole here), because -3 causes the denominator to go to 0 and we all know that having a 0 in the denominator of a fraction is a big no-no!!!
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<h3>Answer: Choice B</h3>
which is the same as x/((x+6)(x-2)) - 3(x-2)/((x+6)(x-2))
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Explanation:
The LCD is (x+6)(x-2) which is the factorization of x^2+4x-12, and that is the denominator of the first fraction. The first fraction has the LCD already. The second fraction does not. It has (x+6) but it is missing (x-2).
We multiply top and bottom of the second fraction by (x-2) to get the second fraction to have the LCD.
turns into
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So,
This is the same as x/((x+6)(x-2)) - 3(x-2)/((x+6)(x-2))
Note the parenthesis around "(x+6)(x-2)"
Instead of x/(x+6)(x-2) you should write x/( (x+6)(x-2) ) to ensure that all of "(x+6)(x-2)" is in the denominator.