If f(x)=3^2 g(x)= 1/6+x find the following and give the domain
(f+g) (x) (3^2) + (1/6+x) x is all real numbers and x is not equal to 6
(f-g) (x) (3^2) - (1/6+x) 3^2 -6-x x is all real numbers and x is not equal to -6
(f x g) (x) I do not understand
(f/g) (x) 3^2/6-x x is all real numbers and x is not equal to 6
To find the sum, difference, product, and quotient of the given functions f(x) and g(x), we can substitute the values of f(x) and g(x) into the respective operations.
1) (f+g)(x) = f(x) + g(x)
Substituting the values, we get:
(f+g)(x) = (3^2) + (1/6+x)
Simplifying, we have:
(f+g)(x) = 9 + (1/6+x)
(f+g)(x) = 9 + (1/6) + x
(f+g)(x) = (9 + 1/6) + x
The domain of (f+g)(x) is all real numbers, except x ≠ 6 since there is a denominator of (6+x).
2) (f-g)(x) = f(x) - g(x)
Substituting the values, we get:
(f-g)(x) = (3^2) - (1/6+x)
Simplifying, we have:
(f-g)(x) = 9 - (1/6+x)
(f-g)(x) = 9 - (1/6) - x
(f-g)(x) = (9 - 1/6) - x
The domain of (f-g)(x) is also all real numbers, except x ≠ -6 since there is a denominator of (6+x).
3) (f x g)(x) = f(x) * g(x)
Substituting the values, we get:
(f x g)(x) = (3^2) * (1/6+x)
Simplifying, we have:
(f x g)(x) = 9 * (1/6+x)
(f x g)(x) = (9/1) * (1/6+x)
(f x g)(x) = 9/6 + 9x
The domain of (f x g)(x) is all real numbers since there are no restrictions.
4) (f/g)(x) = f(x) / g(x)
Substituting the values, we get:
(f/g)(x) = (3^2) / (1/6+x)
Simplifying, we have:
(f/g)(x) = 9 / (1/6+x)
To divide by a fraction, we multiply by its reciprocal:
(f/g)(x) = 9 * (6+x)/1
(f/g)(x) = (9 * (6+x))/1
(f/g)(x) = (54 + 9x)/1
The domain of (f/g)(x) is all real numbers, except x ≠ 6 since there is a denominator of (1/6+x).