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i dont know how to solve this questions without a graphing on a calculator but i can't use calculator for this question pls help

i dont know how to solve this questions without a graphing on a calculator but i cant use calculator for this question pls help class=

Respuesta :

Averysmartperson

Answer:

D

Step-by-step explanation:

When the denominator is equal to 0, the function doesn't exist. So factor the bottom, luckily its already partially factored.

(x^2 + 8x + 12) = (x+2)(x+6)

So now the denominator is (x+2)(x+6)(x-3)

The function doesn't exist when x = -2, -6, 3

Now we know there are 3 total discontinuities.

A and B can be eliminated

Since there is an (x+2) on the top and the bottom, they don't affect the shape of the function, but only causes a removable discontinuity when x = -2.

However, (x+6) and (x-3) are only on the bottom, so they DO change the shape, so they are non-removable

The answer therefore is 1 removable and 2 non-removable; D

mirai123

Answer:

We have 1 removable discontinuity and 2 non-removable discontinuities

Step-by-step explanation:

Note that

[tex]x^2+8x+12 = (x+2)(x+6)[/tex]

Initially, we have discontinuities at

[tex]x = -2; x = -6; x = 3[/tex]

Considering

[tex]f(x)=\dfrac{(x+2)(x-6)}{(x^2+8x+12)(x-3)}[/tex]

But

[tex]\dfrac{(x+2)(x-6)}{(x^2+8x+12)(x-3)}= \dfrac{(x+2)(x-6)}{ (x+2)(x+6)(x-3)}= \dfrac{(x-6)}{ (x+6)(x-3)}[/tex]

We have now two discontinuities at

[tex]x = -6; x = 3[/tex]

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