Answer
Domain f(x) = x ≥ -1
Range f(x) = y ≥ 0
Domain g(x) = x ≥ 1
Range g(x) = y ≥ 0
Domain f(x) + g(x) = x ≥ 1 (common domain)
Range f(x) + g(x) = y ≥ $\sqrt2$
Domain f(x)g(x) = x ≥ 1 (common domain)
Range f(x)g(x) = y ≥ 0
Work Step by Step
For the domain of f(x) the answers of the square root need to be real numbers. So $x+1≥0=x≥-1$. Thus domain f(x) = x≥-1.
The range of a square root function is y≥0.
For the domain of f(x) the answers of the square root need to be real numbers. So $x−1≥0=x≥1$. Thus domain g(x) = x≥1.
The range of a square root function is y≥0.
(f + g)(x) = $\sqrt{x+1} + \sqrt {x−1}$. The domain for (f + g)(x) is the common domain of f(x) and g(x). Thus the domain of (f + g)(x) = x≥1.
(f + g)(1) =$\sqrt{1+1} + \sqrt {1−1} = \sqrt{2} + \sqrt {0} = \sqrt{2} + 0 = \sqrt{2}$. So the range of (f+g)(x) = $y≥\sqrt2$.
(f×g)(x)=f(x)g(x)=$\sqrt{x+1} \times \sqrt{x-1} = \sqrt{(x+1)(x-1)} = \sqrt{x^2-1}$. The domain for f(x)g(x) is the common domain of f(x) and g(x). Thus the domain of f(x)g(x) = x≥1. f(1)g(1) = $\sqrt{1^2-1} = \sqrt {1−1} = \sqrt{0} = 0$. So the range of f(x)g(x) = y≥0.
