đạo hàm

N

noinhobinhyen

$f'(x) = x^2-4x+m$

1. $f'(x) > 0 \forall x > 0 (1)$

$\Leftrightarrow x^2-4x > -m \forall x > 0$

do $x^2-4x \geq -4 \forall x>0$

nên (1) đúng $\forall x > 0 \Leftrightarrow -m < -4 \Leftrightarrow m>4$

2.

$f'(x) < 0 \forall x \in (0;2) (2)$

$\Leftrightarrow x^2-4x < -m \forall x \in (0;2) $

Do $x^2-4x < 0\forall x \in (0;2)$

nên (2) đúng $\forall x \in (0;2) \Leftrightarrow -m \geq 0 \Leftrightarrow m \leq 0$
 
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