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