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

[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, ∞).