Hàm số y=x4−4x3−5 nhận điểm
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
$y' = 4x^3 - 12x^2 = 4x^2(x-3)$
$y' = 0 \Leftrightarrow \begin{cases} x = 0 \\ x = 3 \end{cases}$
$y'' = 12x^2 - 24x$
$y''(0) = 0$ (không kết luận được)
$y''(3) = 12 \cdot 3^2 - 24 \cdot 3 = 36 > 0$, suy ra $x=3$ là điểm cực tiểu.
Xét dấu $y'$:
- Khi $x < 0$, $y' < 0$
- Khi $0 < x < 3$, $y' < 0$
- Khi $x > 3$, $y' > 0$
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