Find the range of the following $f(x) = \sqrt{x^2 -3x+2}$
Answer & explanation
Correct answer: option 1
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, ∞).