Which of the following is a rational expression whose domain has a restriction of x


=−1?
A.
x
3

x+1

B.
2x+2
x



C.
6x+6
x
2
+6x+5

D.
x
6x
3
+60

3 answers

To determine which rational expression has a restriction of \( x \neq -1 \) in its domain, we need to identify expressions where the denominator equals zero when \( x = -1 \).

Let's analyze each option:

A.
\[ \frac{x^3}{x + 1} \] The denominator is \( x + 1 \). Setting this equal to zero: \[ x + 1 = 0 \implies x = -1 \] Thus, this expression has a restriction of \( x \neq -1 \).

B.
\[ \frac{2x + 2}{x} \] The denominator is \( x \). Setting this equal to zero: \[ x = 0 \] This expression has a restriction of \( x \neq 0 \).

C.
\[ \frac{6x + 6}{x^2 + 6x + 5} \] The denominator is \( x^2 + 6x + 5 \). Factoring the quadratic: \[ x^2 + 6x + 5 = (x + 1)(x + 5) \] Setting this equal to zero: \[ (x + 1)(x + 5) = 0 \implies x = -1 \text{ or } x = -5 \] This expression has a restriction of \( x \neq -1 \) and \( x \neq -5 \).

D.
\[ \frac{x}{6x^3 + 60} \] The denominator is \( 6x^3 + 60 \). Setting this equal to zero: \[ 6x^3 + 60 = 0 \implies 6x^3 = -60 \implies x^3 = -10 \implies x = -\sqrt[3]{10} \] Thus, there is no restriction at \( x = -1 \).

Conclusion:

Both options A and C have restrictions of \( x \neq -1 \) in their domain. However, the question looks for only one option. Since the explicit mention of the restriction is required and since both A and C fulfill that requirement, the best fit for the answer that represents a simplistic understanding would be option A.

Therefore, the answer is: A. \( \frac{x^3}{x + 1} \).

Are you sure

Let's reevaluate the options to confirm the rational expressions and their respective restrictions clearly:

Options:

A.
\[ \frac{x^3}{x + 1} \]
Denominator: \( x + 1 \)
Setting \( x + 1 = 0 \):
\[ x = -1
\]
Restriction: \( x \neq -1 \)


B.
\[ \frac{2x + 2}{x} \]
Denominator: \( x \)
Setting \( x = 0 \):
\[ x = 0
\]
Restriction: \( x \neq 0 \)


C.
\[ \frac{6x + 6}{x^2 + 6x + 5} \]
Denominator: \( x^2 + 6x + 5 \)
Factoring gives:
\[ x^2 + 6x + 5 = (x + 1)(x + 5)
\]
Setting \( (x + 1)(x + 5) = 0 \):
\[ x = -1 \text{ or } x = -5
\]
Restrictions: \( x \neq -1 \) and \( x \neq -5 \)


D.
\[ \frac{x}{6x^3 + 60} \]
Denominator: \( 6x^3 + 60 \)
Setting \( 6x^3 + 60 = 0 \):
\[ 6x^3 = -60 \implies x^3 = -10 \implies x = -\sqrt[3]{10}
\]
Restriction: \( x \neq -\sqrt[3]{10} \)


Summary of Restrictions:

  • A: \( x \neq -1 \)
  • B: \( x \neq 0 \)
  • C: \( x \neq -1, -5 \)
  • D: \( x \neq -\sqrt[3]{10} \)

Conclusion:

Both A and C contain the restriction \( x \neq -1 \), but since the question asks for a rational expression with this specific restriction and no other restrictions, the most straightforward choice is:

A. \( \frac{x^3}{x + 1} \), since it directly restricts \( x \) only at \(-1\) and does not impose any other restrictions.

Thank you for your patience, and I hope this clears up any confusion!