Find $f+g$, $f-g$, $fg$, and $\frac{f}{g}$. Determine the domain for each function.
$f\left(x\right)=6x^2-x-1$, $g\left(x\right)=x-1$

Verified step by step guidance

1. To find \( (f+g)(x) \), add the functions \( f(x) \) and \( g(x) \): \( (f+g)(x) = f(x) + g(x) = (6x^2 - x - 1) + (x - 1) \). Combine like terms to simplify.
2. To find \( (f-g)(x) \), subtract \( g(x) \) from \( f(x) \): \( (f-g)(x) = f(x) - g(x) = (6x^2 - x - 1) - (x - 1) \). Distribute the negative sign and combine like terms.
3. To find \( (fg)(x) \), multiply the functions \( f(x) \) and \( g(x) \): \( (fg)(x) = f(x) \cdot g(x) = (6x^2 - x - 1)(x - 1) \). Use the distributive property (FOIL method) to expand the expression.
4. To find \( \left(\frac{f}{g}\right)(x) \), divide \( f(x) \) by \( g(x) \): \( \left(\frac{f}{g}\right)(x) = \frac{6x^2 - x - 1}{x - 1} \). Simplify the expression if possible by factoring the numerator and checking for common factors with the denominator.
5. Determine the domain for each function: For \( f+g \) and \( f-g \), the domain is all real numbers since there are no restrictions. For \( fg \), the domain is also all real numbers. For \( \frac{f}{g} \), the domain excludes \( x = 1 \) because it makes the denominator zero, which is undefined.