For the following pair of functions, f(x)-x+2, g(x)-√x-1
find each of the following and state its domain:
a.) find (f+g)(x)
b.)find (f-g) (x)
c.) find (fg)(x)
d.) find (ff)(x)
e.)Find (f/g)(x)
looks like a typo in both functions.
is it f(x) = x+2 or f(x) = -x+2 ?
same for the other one.
To find the given expressions and their domains, we'll start with the basic operations of addition, subtraction, multiplication, and division.
a) To find (f+g)(x), we need to add the functions f(x) and g(x):
(f+g)(x) = f(x) + g(x)
= (x+2) + (√x-1)
= x + 2 + √x - 1
The domain of (f+g)(x) will be the intersection of the domains of f(x) and g(x).
For f(x) = x+2, the domain is all real numbers since there are no restrictions.
For g(x) = √x - 1, the domain is x ≥ 0 (non-negative real numbers) because we cannot take the square root of a negative number.
Therefore, the domain of (f+g)(x) will be x ≥ 0.
b) To find (f-g)(x), we need to subtract g(x) from f(x):
(f-g)(x) = f(x) - g(x)
= (x+2) - (√x-1)
= x + 2 - √x + 1
The domain of (f-g)(x) will also be x ≥ 0 because we have the same functions involved as in (a).
c) To find (fg)(x), we need to multiply f(x) and g(x):
(fg)(x) = f(x) * g(x)
= (x+2) * (√x-1)
= x√x - x + 2√x - 2
The domain of (fg)(x) will be x ≥ 0 as we have the same functions involved again.
d) To find (ff)(x), we need to use the function f(x) as input for itself:
(ff)(x) = f(f(x))
= f(x+2)
= (x+2) + 2 = x + 4
The domain of (ff)(x) will also be all real numbers since there are no restrictions.
e) To find (f/g)(x), we need to divide f(x) by g(x):
(f/g)(x) = f(x) / g(x)
= (x+2) / (√x-1)
The domain of (f/g)(x) will be x > 1, as we have the square root involved, and the denominator cannot be equal to zero.
In summary, we have:
a) (f+g)(x) = x + 2 + √x - 1; Domain: x ≥ 0
b) (f-g)(x) = x + 2 - √x + 1; Domain: x ≥ 0
c) (fg)(x) = x√x - x + 2√x - 2; Domain: x ≥ 0
d) (ff)(x) = x + 4; Domain: All real numbers
e) (f/g)(x) = (x+2) / (√x-1); Domain: x > 1