Đáp án:
$b)\underset{(-2;+\infty)}{max \ } y=1 \Leftrightarrow x=-1\\ d)\underset{(1;+\infty)}{min \ } y=-\dfrac{1}{4} \Leftrightarrow x=2\\ f)\underset{(-3;+\infty)}{max \ } y=1 \Leftrightarrow x=-2.$
Giải thích các bước giải:
$b)y=\dfrac{-x^2-x+1}{x+2}, x>-2\\ y'=\dfrac{(-x^2-x+1)'(x+2)- (-x^2-x+1)(x+2)'}{(x+2)^2}\\ =\dfrac{(-2x-1)(x+2)- (-x^2-x+1)}{(x+2)^2}\\ =\dfrac{-x^2 - 4 x - 3}{(x+2)^2}\\ y'=0 \Leftrightarrow x=-1,x=-3\\ BBT:$
\begin{array}{|c|ccccccccc|} \hline x&-2&&-1&&\infty\\\hline y'&&+&0&-&\\\hline &&&1\\y&&\nearrow&&\searrow&\\&-\infty&&&&-\infty\\\hline\end{array}
Dựa vào $BBT \Rightarrow \underset{(-2;+\infty)}{max \ } y=1 \Leftrightarrow x=-1$
$d)y=\dfrac{-x}{x^2+4}, x>1\\ y'=\dfrac{-x'(x^2+4)+x(x^2+4)'}{(x^2+4)^2}\\ =\dfrac{-(x^2+4)+x.2x}{(x^2+4)^2}\\ =\dfrac{x^2 - 4}{(x^2+4)^2}\\ y'=0 \Leftrightarrow x=\pm 2\\ BBT:$
\begin{array}{|c|ccccccccc|} \hline x&1&&2&&\infty\\\hline y'&&-&0&+&\\\hline &-\dfrac{1}{5}&&&&0\\y&&\searrow&&\nearrow&\\&&&-\dfrac{1}{4}\\\hline\end{array}
Dựa vào $BBT \Rightarrow \underset{(1;+\infty)}{min \ } y=-\dfrac{1}{4} \Leftrightarrow x=2$
$f)y=-x-\dfrac{1}{x+3}, x>-3\\ y'=-1+\dfrac{1}{(x+3)^2}\\ =\dfrac{-(x+3)^2+1}{(x+3)^2}\\ =\dfrac{-x^2 - 6 x - 8}{(x+3)^2}\\ y'=0 \Leftrightarrow x=-4;x=-2\\ BBT:$
\begin{array}{|c|ccccccccc|} \hline x&-3&&-2&&\infty\\\hline y'&&+&0&-&\\\hline &&&1\\y&&\nearrow&&\searrow&\\&-\infty&&&&-\infty\\\hline\end{array}
Dựa vào $BBT \Rightarrow \underset{(-3;+\infty)}{max \ } y=1 \Leftrightarrow x=-2.$
