6. Diketahui fungsi-fungsi f dan g masing-masing ditentukan dengan f(x)=x^2+x dan g(x)=(2)/(x+3) Tentukan fungsi-fungsi berikut beserta dengan domainnya! a. (f+g)(x) c. (fcdot g)(x) b. (f-g)(x) d. (f/g)(x)

**a. (f+g)(x)**(f+g)(x) = f(x) + g(x) = (x² + x) + (2/(x+3)) = x² + x + 2/(x+3)**Domain:** The domain of (f+g)(x) is the intersection of the domains of f(x) and g(x). f(x) = x² + x has a domain of all real numbers (-∞, ∞). g(x) = 2/(x+3) is undefined when the denominator is zero, i.e., when x = -3. Therefore, the domain of (f+g)(x) is (-∞, -3) U (-3, ∞).**b. (f-g)(x)**(f-g)(x) = f(x) - g(x) = (x² + x) - (2/(x+3)) = x² + x - 2/(x+3)**Domain:** The domain is the same as in part (a), because the only restriction comes from the denominator of g(x). Therefore, the domain of (f-g)(x) is (-∞, -3) U (-3, ∞).**c. (f·g)(x)**(f·g)(x) = f(x) * g(x) = (x² + x) * (2/(x+3)) = 2x(x+1)/(x+3)**Domain:** Again, the denominator cannot be zero, so x ≠ -3. Also, we must consider if there are any restrictions from the numerator. The numerator is 2x(x+1), which is defined for all real numbers. Therefore, the domain of (f·g)(x) is (-∞, -3) U (-3, ∞).**d. (f/g)(x)**(f/g)(x) = f(x) / g(x) = (x² + x) / (2/(x+3)) = (x² + x)(x+3)/2 = x(x+1)(x+3)/2**Domain:** The expression simplifies to a polynomial, but we must still consider the original g(x) in the denominator. The original function (f/g)(x) is undefined when g(x) = 0, which never happens (since the numerator of g(x) is a constant). However, it's also undefined when the denominator of g(x) is zero (x = -3). Therefore, the domain of (f/g)(x) is (-∞, -3) U (-3, ∞).In summary:a. (f+g)(x) = x² + x + 2/(x+3); Domain: (-∞, -3) U (-3, ∞)b. (f-g)(x) = x² + x - 2/(x+3); Domain: (-∞, -3) U (-3, ∞)c. (f·g)(x) = 2x(x+1)/(x+3); Domain: (-∞, -3) U (-3, ∞)d. (f/g)(x) = x(x+1)(x+3)/2; Domain: (-∞, -3) U (-3, ∞)