To determine which rational expression has a domain restriction of \( x \neq -1 \), we need to analyze each option to see if substituting \( x = -1 \) causes any discontinuities (like division by zero) or is otherwise undefined.
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\( \frac{x+1}{x^3} \):
- The denominator \( x^3 \) is zero when \( x = 0 \) (which is a restriction), but not at \( x = -1 \).
- Domain: all \( x \) except \( x = 0 \).
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\( \frac{x^2 + 6x + 5}{6x + 6} \):
- The denominator \( 6x + 6 = 6(x + 1) \) is zero when \( x = -1 \).
- Domain: all \( x \) except \( x = -1 \).
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\( \frac{\sqrt{x}}{2x + 2} \):
- The denominator \( 2x + 2 = 2(x + 1) \) is zero when \( x = -1 \) and the square root is defined only for \( x \geq 0 \).
- Domain: \( x \geq 0\) except \( x = -1\), but since \( -1 \) is not in the domain anyway, it’s valid.
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\( 6x^3 + 60x \):
- This is a polynomial and is defined for all \( x \). There are no restrictions.
- Domain: all \( x \).
Out of these options, the only rational expression whose domain has a restriction of \( x \neq -1 \) is:
\( \frac{x^2 + 6x + 5}{6x + 6} \).