Question

Which of the following functions have only one point that is not in the domain? Select all that apply. (There are 2 answers)

A. f(x)=x^2-6x-8
B. f(x)=3/x^2+2x+1
C. f(x)=sqrt 2x-1
D. f(x)=1/x^2+5x+4
E. f(x)=5/3+x

Answer 1

Answer:

  B.  f(x) = 3/(x^2 +2x +1)

  E.  f(x) = 5/(3+x)

Step-by-step explanation:

The above functions have denominators that are zero at one point. For B, that point is x=-1; for E, that point is x=-3. When the denominator is zero, the function is undefined, so the corresponding x-value must be excluded from the domain.

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As written, choice D also qualifies. This is because the function written in the answer list here is ...

$f(x)=\dfrac{1}{x^2}+5x+4$

It will be undefined for x=0.

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Comment on writing rational functions

When the division symbol (/ or ÷) is used with a denominator that is a product or sum, that product or sum must be enclosed in parentheses. When the expression is typeset, the division bar acts as a grouping symbol. When the expression is written in text, a grouping symbol, such as parentheses, must be added.

Answer 2

• B. $f(x) = \frac{3}{x^2 + 2x + 1}$
• E. $f(x) = \frac{5}{3 + x}$

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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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Item 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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Item b:

$f(x) = \frac{3}{x^2+2x+1}$

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

The denominator is $x^2 + 2x + 1$, and it can be factored as a perfect square trinomial:

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

It is zero when:

$(x + 1)^2 = 0$

$\sqrt{(x + 1)^2} = \sqrt{0}$

$x + 1 = 0$

$x = -1$

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

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Item c:

$f(x) = \sqrt{2x-1}$

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

$2x - 1 \geq 0$

$2x \geq 1$

$x \geq \frac{1}{2}$

Thus, there are infinite values $x < \frac{1}{2}$ not in the domain.

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Item d:

$f(x) = \frac{1}{x^2+5x+4}$

Another fraction, and the denominator can be factored as:

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

Which is zero at:

$x + 4 = 0 \rightarrow x = -4$

$x + 1 = 0 \rightarrow x = -1$

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

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Item e:

$f(x) = \frac{5}{3+x}$

Fraction, the denominator cannot be 0. So

$3 + x \neq = 0 \rightarrow x \neq 3$

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

Thus, options B and E are selected.

A similar question is given at https://brainly.com/question/24210366