Đáp án:
Giải thích các bước giải:
$a)\\y' = 3x^2 + 6x\\TXD:D = \mathbb{R}\\y' = 0\\\Leftrightarrow 3x^2 + 6x = 0\\\Leftrightarrow\begin{cases} x=0\\ x=-2\end{cases}\\\text{ta có BBT}\\\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}&-2\text{}&\text{}&0\text{}&\text{$$}&\text{}&\text{}+\infty\\\hline\text{$y'$}&\text{}&+\text{}&\text{}&0\text{}&-\text{}&0&&+& \\\hline \text{$y$}&\text{}&\text{}&3\text{}&\text{}&\text{}&\text{}&-\infty\text{}&\text{}&\text{}\\ &\text{}&\text{$\nearrow$}&\text{}&\text{}\searrow&\text{}&\text{}\nearrow&\text{}&\text{$$}\\&-\infty\text{$$}&\text{}&\text{}&\text{}&-1\text{}&\text{}&\text{}&&\\\hline \end{array}\\\text{đồng biến trên }\begin{cases} (-\infty ; - 5)\\(0 ; +\infty) \end{cases}\\\text{nghịch biến trên }(-2;0)\\b)\\y'=-3x^2 + 3\\TXD : D = \mathbb R\\y' = 0\\\Rightarrow x = \pm 1\\\text{ta có BBT }\\\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}-1&\text{}&\text{}1&\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}-&\text{0}&\text{}+&\text{0}&\text{}-&\\\hline \text{$y$}&\text{}+\infty&\text{}&\text{}&\text{}&\text{}-2&\text{}\\&\text{}&\text{$\searrow$}&\text{}&\text{}\nearrow&\text{}&\text{}\searrow\\&\text{$$}&\text{}&\text{}-6&\text{}&\text{}&\text{}&\text{}-\infty\\\hline \end{array}\\\text{đồng biến trên }(-1;1)\\\text{nghịch biến trên }\begin{cases} (-\infty;-1)\\(1 ; + \infty) \end{cases}\\c)\\y' = 4x^3 - 4x\\y' = 0\\\Rightarrow \begin{cases} x = 0\\x = \pm 1\end{cases}\\\text{ta có BBT :}\\\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}-1&\text{}&\text{}0&\text{}&1\text{$$}&&+\infty\\\hline \text{$y'$}&\text{}&\text{}-&\text{}0&\text{}+&0\text{}&-\text{}&0&+&\\\hline \text{$y$}&\text{}+\infty&\text{}&\text{}&\text{}&\text{}-1&\text{}&&&+\infty\\&\text{}&\text{$\searrow$}&\text{}&\text{}\nearrow&\text{}&\text{}\searrow&&\nearrow\\&\text{$$}&\text{}&\text{}-2&\text{}&\text{}&\text{}&\text{}-2\\\hline \end{array} \\\text{đồng biến trên}\begin{cases}(0;-1)\\(1;+\infty)\end{cases}\\\text{nghịch biến trên}\begin{cases}(-\infty;-1)\\(0;1)\end{cases}\\d)\\y' = 4x^3 + 4x\\TXD:D = \mathbb R\\y' = 0\\\Rightarrow \begin{cases} x = 0\\x = \pm 1 \end{cases}\\\text{ta có BBT}\\\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}-1&\text{}&\text{}0&\text{}&1\text{$$}&&+\infty&\\\hline \text{$y'$}&\text{}&\text{}+&\text{0}&\text{}-&\text{0}&\text{}+&0&-&&\\\hline \text{$y$}&\text{}&\text{}&15\text{}&\text{}&\text{}&\text{}&15\\&\text{}&\text{$\nearrow$}&\text{}&\text{}\searrow&\text{}&\text{}\nearrow& &\searrow&\\&\text{$$}-\infty&\text{}&\text{}&\text{}&4\text{}&\text{}&\text{}& &-\infty&\\\hline \end{array}\\\text{đồng biếng trên }\begin{cases} (-\infty ; -1)\\(0;1) \end{cases}\\\text{nghịch biến trên }\begin{cases} (-1;0)\\(1 ; +\infty)\end{cases}\\e)\\TXD : D = \mathbb{R}\text{\\}\{-3\}\\y' = \dfrac{7}{(x + 3)^2} > 0\text{ }\forall x + -3\\\text{ta có BBT}\\\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}&\text{}3&\text{}&\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}&\text{+}&\text{}∥&\text{+}&\text{}&\\\hline \text{$y$}&\text{}&\text{}&\text{}+\infty&\text{}∥&\text{}&\text{}&2\\&\text{}&\text{$\nearrow$}&\text{}&\text{}∥&\text{}&\text{}\nearrow\\&2\text{$$}&\text{}&\text{}&∥\text{}&-\infty\text{}&\text{}&\text{}\\\hline \end{array}\\\text{đồng biến trên }\begin{cases} (-\infty ; - 3)\\(-3 ; +\infty) \end{cases}\\\mathscr{X_{\textit{in hay nhất}}}$
