Đáp án:
Giải thích các bước giải:
$1)\\y = x^3 - 9x^2 + 24x - 2\\ \text{TXD: D = $\mathbb R$}\\ y' = 3x^2 - 18x + 24\\ y' = 0 \\\Leftrightarrow \left[\begin{array}{l}x = 2\\x = 4\end{array}\right.\\ \text{ta có BBT :}\\\begin{array}{c|cc} \text{$x$}&\text{$-\infty$}&\text{}&\text{}2&\text{}&\text{}4&\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}+&\text{0}&\text{}-&\text{0}&\text{}+&\\\hline \text{$y$}&\text{}&\text{}&18\text{}&\text{}&\text{}&\text{}&+\infty\\&\text{}&\text{$\nearrow$}&\text{}&\text{}\searrow&\text{}&\text{}\nearrow\\&\text{$$}-\infty&\text{}&\text{}&\text{}&\text{}14&\text{}&\text{} \end{array}\\ \text{đồng biến trên }\begin{cases} (-\infty;2) \\(4;+\infty)\end{cases}\\ \text{nghịch biến trên } (2;4)\\2)\\y = x^3 - 3x^2 + 3x + 5\\ \text{TXD : D = $\mathbb R$}\\ y' \ = 3x^2 - 6x + 3 \\\quad = 3(x-1)^2 \ge 0\\\text{Vậy đồng biến trên }\mathbb R\\3)\\\quad y = -x^3 - 3x - 2\\\text{TXD : D = $\mathbb R$}\\ y ' \ = -3x^2 - 3 \\\quad= -3(x^2 + 1)\le 0\\\text{Vậy nghịch biến trên }\mathbb R$
