Đáp án:

$b)\underset{(-\infty;2)}{min \ } y=-5 \Leftrightarrow x=1\\ d)\underset{(-\infty;0)}{max \ } y=\dfrac{1}{6} \Leftrightarrow x=-6\\ f)\underset{\left(-\infty;\tfrac{3}{8}\right)}{min \ } y=\dfrac{1}{4} \Leftrightarrow x=\dfrac{1}{8}.$

Giải thích các bước giải:

$b)y=\dfrac{-x^2-3x+9}{x-2}, x<2\\ y'=\dfrac{(-x^2-3x+9)'(x-2) - (-x^2-3x+9)(x-2)'}{(x-2)^2}\\ =\dfrac{(-2x-3)(x-2) - (-x^2-3x+9)}{(x-2)^2}\\ =\dfrac{-x^2 + 4 x - 3}{(x-2)^2}\\ y=0 \Leftrightarrow x=1;x=3\\ BBT:$

\begin{array}{|c|ccccccccc|} \hline x&-\infty&&1&&2\\\hline y'&&-&0&+&\\\hline &+\infty&&&&+\infty\\y&&\searrow&&\nearrow&\\&&&-5\\\hline\end{array}

Dựa vào $BBT\Rightarrow \underset{(-\infty;2)}{min \ } y=-5 \Leftrightarrow x=1$

$d)y=\dfrac{-2x}{x^2+36}, x<0\\ y'=\dfrac{(-2x)'(x^2+36)+2x(x^2+36)'}{(x^2+36)^2}\\ =\dfrac{-2(x^2+36)+2x.2x}{(x^2+36)^2}\\ =\dfrac{2 x^2 - 72}{(x^2+36)^2}\\ y=0 \Leftrightarrow x=\pm 6\\ BBT:$

\begin{array}{|c|ccccccccc|} \hline x&-\infty&&-6&&0\\\hline y'&&+&0&-&\\\hline &&&\dfrac{1}{6}\\y&&\nearrow&&\searrow&\\&0&&&&0\\\hline\end{array}

Dựa vào $BBT\Rightarrow \underset{(-\infty;0)}{max \ } y=\dfrac{1}{6} \Leftrightarrow x=-6$

$f)y=-2x-\dfrac{1}{8x-3}, x<\dfrac{3}{8}\\ y'=-2+\dfrac{8}{(8x-3)^2}\\ =\dfrac{8-2(8x-3)^2}{(8x-3)^2}\\ =\dfrac{-128 x^2 + 96 x - 10}{(8x-3)^2}\\ y=0 \Leftrightarrow x=\dfrac{1}{8}; x=\dfrac{5}{8}\\ BBT:$

\begin{array}{|c|ccccccccc|} \hline x&-\infty&&\dfrac{1}{8}&&\dfrac{3}{8}\\\hline y'&&-&0&+&\\\hline &+\infty&&&&+\infty\\y&&\searrow&&\nearrow&\\&&&\dfrac{1}{4}\\\hline\end{array}

Dựa vào $BBT\Rightarrow \underset{\left(-\infty;\tfrac{3}{8}\right)}{min \ } y=\dfrac{1}{4} \Leftrightarrow x=\dfrac{1}{8}.$