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If $f(x) = x^2 - 4x + 3$, the **interval** on which $f$ is decreasing is:
A$x < 2$
B$x > 2$
CAll $x$
D$x = 2$
Answer & Solution
Correct answer: A. $x < 2$
$f'(x) = 2x - 4 < 0$ ⇔ $x < 2$. So $f$ is decreasing on $(-\infty, 2)$ and increasing on $(2, \infty)$. Vertex/minimum at $x = 2$.
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