Question

HELP PLEASE pre-calc

which of the following functions have only one point that is not in the domain? select all that apply

Answer 1

1. B:
   `f(x)=(3)/(x^2+2x+1)`
2. E:
   `f(x)=(5)/(3+x)`

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The domain of a function are the values that the input can assume, in this case, the possible values of x.

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Answer choice A:

`f(x)=x^2-6x+8`

It is a simple quadratic function, with no impediments, so the domain is all real numbers, that is, there are no points that are not in the domain.

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Answer choice B:

`f(x)=(3)/(x^2+2x+1)`

It is a fraction, so the impediment is that the denominator cannot be zero.

The denominator is , and it can be factored as a perfect square trinomial:

`x^2+2x+1=(x+1)^2`

It is zero when:

`(x+1)^2=0`

`√((x+1)^2) =√(0)`

`x+1=0`

`x=-1`

Thus, it has only one point, x = -1, that is not in the domain.

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Answer choice C:

`f(x)=√(2x-1)`

The square root only exists for non-negative values, thus, the domain is:

`2x-1\geq 0`

`2x\geq 1`

`x\geq (1)/(2)`

Thus, there are infinite values
`x < (1)/(2)` not in the domain.

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Answer choice D:

`f(x)=(1)/(x^2+5x+4)`

Another fraction, and the denominator can be factored as:

`x^2+5x+4=(x+4)(x+1)`

Which is zero at:

`x+4=0` →
`x=-4`

`x+1=0` →
`x=-1`

Thus, it has two points, x = -4 and x = -1, not in the domain.

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Answer choice E:

`f(x)=(5)/(3+x)`

Fraction, the denominator cannot be 0. So

`3+x\\eq 0` →
`x\\eq 3`

There is one point, x = -3, not in the domain.

Thus, options B and E are selected.

Answer 2

the second choice, and the last choice are correct.

the first one, x can be any number.
the second one, x can be any number but not -1(the denominator cannot be 0)
the third one, x has to be ≥ 1/2, because 2x-1 cannot be negative.
the fourth one, x can be any number but not -1 or -4
the last one, x can be any number but not -3