## Thomas' Calculus 13th Edition

Domain: x ∈ ($-\infty$,0] ∪ [3,$\infty$) Range: y ∈ (-$\infty$,$\infty$)
Domain: According to the question, f(x)= √(x^2 - 3x) Now, for f(x) to be real (x^2 - 3x) $\geq$ 0 ⇒ (x)(x-3)$\geq$0 ⇒ x$\geq$0 AND x$\geq$3 [ Since positive number X postive number = postive] OR x$\leq$0 AND x$\leq$3 [ Since negative number X negative number = positive number] ⇒ x$\geq$3 OR x$\leq$0 Therefore, x ∈ ($-\infty$,0] ∪ [3,$\infty$) Range: Since f(x)= √(x^2 - 3x) is any real number, the range of f(x) is (-$\infty$,$\infty$)