Find the range of the following $f(x) = \sqrt{x^2 -3x+2}$

[0, ∞) [-∞, ∞) [-∞, 0) [-∞, 1)

[0, ∞)

Clearly f(x) is defined if $x^2 - 3x+2≥0$ $⇒ (x-1) (x-2) ≥0$ $⇒ x∈ (-∞,1]∪[2, ∞)$ Now, $f(x) = \sqrt{x^2 -3x+2}$ $=\sqrt{(x-\frac{3}{2})^2+2-\frac{9}{4}}$ $=\sqrt{(x-\frac{3}{2})^2-\frac{1}{4}}$ Clearly, $(x-\frac{3}{2})^2-\frac{1}{4}≥0$ $∴\sqrt{(x-\frac{3}{2})^2-\frac{1}{4}}≥0$ Therefore, range is [0, ∞).