Đáp án:
$a)\underset{(-1+\infty)}{min \ } y=3 \Leftrightarrow x=0\\ b)\underset{(0;+\infty)}{max \ } y=\dfrac{1}{3} \Leftrightarrow x=3\\ c)\underset{(2;+\infty)}{min \ } y=4 \Leftrightarrow x=3.$
Giải thích các bước giải:
$a)y=\dfrac{x^2+3x+3}{x+1}, x>-1\\ y'=\dfrac{(x^2+3x+3)'(x+1) -(x^2+3x+3)(x+1)'}{(x+1)^2}\\ =\dfrac{(2x+3)(x+1) -(x^2+3x+3)}{(x+1)^2}\\ =\dfrac{x^2 + 2 x}{(x+1)^2}\\ y'=0 \Leftrightarrow x=0,x=-2\\ BBT:$
\begin{array}{|c|ccccccccc|} \hline x&-1&&0&&\infty\\\hline y'&&-&0&+&\\\hline &+\infty&&&&+\infty\\y&&\searrow&&\nearrow&\\&&&3\\\hline\end{array}
Dựa vào $BBT \Rightarrow \underset{(-1+\infty)}{min \ } y=3 \Leftrightarrow x=0$
$b)y=\dfrac{2x}{x^2+9},x>0\\ y'=\dfrac{(2x)'(x^2+9)-2x(x^2+9)'}{(x^2+9)^2}\\ =\dfrac{2(x^2+9)-2x.2x}{(x^2+9)^2}\\ =\dfrac{18 - 2 x^2}{(x^2+9)^2}\\ y'=0 \Leftrightarrow x= \pm 3\\ BBT:$
\begin{array}{|c|ccccccccc|} \hline x&0&&3&&\infty\\\hline y'&&+&0&-&\\\hline &&&\dfrac{1}{3}\\y&&\nearrow&&\searrow&\\&0&&&&0\\\hline\end{array}
Dựa vào $BBT \Rightarrow \underset{(0;+\infty)}{max \ } y=\dfrac{1}{3} \Leftrightarrow x=3$
$c)y=x+\dfrac{1}{x-2},x>2\\ y'=1-\dfrac{1}{(x-2)^2}\\ =\dfrac{(x-2)^2-1}{(x-2)^2}\\ =\dfrac{x^2 - 4 x + 3}{(x-2)^2}\\ y'=0 \Leftrightarrow x=1,x=3\\ BBT:$
\begin{array}{|c|ccccccccc|} \hline x&2&&3&&\infty\\\hline y'&&-&0&+&\\\hline &+\infty&&&&+\infty\\y&&\searrow&&\nearrow&\\&&&4\\\hline\end{array}
Dựa vào $BBT \Rightarrow \underset{(2;+\infty)}{min \ } y=4 \Leftrightarrow x=3.$
