Đáp án:
$b)\underset{(-\infty;5)}{min \ } y=-1 \Leftrightarrow x=4\\ d)\underset{(-\infty;1)}{max \ } y=\dfrac{1}{2} \Leftrightarrow x=-1\\ f)\underset{(-\infty;-5)}{min \ } y=7 \Leftrightarrow x=-5.$
Giải thích các bước giải:
$b)y=\dfrac{-x^2+7x-11}{x-5}, x<5\\ y'=\dfrac{(-x^2+7x-11)'(x-5)-(-x^2+7x-11)(x-5)'}{(x-5)^2}\\ =\dfrac{(-2x+7)(x-5)-(-x^2+7x-11)}{(x-5)^2}\\ =\dfrac{-x^2 + 10 x - 24}{(x-5)^2}\\ y'=0 \Leftrightarrow x=4,x=6\\ BBT:$
\begin{array}{|c|ccccccccc|} \hline x&-\infty&&4&&5\\\hline y'&&-&0&+&\\\hline &+\infty&&&&+\infty\\y&&\searrow&&\nearrow&\\&&&-1\\\hline\end{array}
Dựa vào $BBT \Rightarrow \underset{(-\infty;5)}{min \ } y=-1 \Leftrightarrow x=4$
$d)y=\dfrac{-x}{x^2+1},x<1\\ y'=\dfrac{-x'(x^2+1)+x(x^2+1)'}{(x^2+1)^2}\\ =\dfrac{-(x^2+1)+x.2x}{(x^2+1)^2}\\ =\dfrac{x^2-1}{(x^2+1)^2}\\ y'=0 \Leftrightarrow x=\pm 1\\ BBT:$
\begin{array}{|c|ccccccccc|} \hline x&-\infty&&-1&&1\\\hline y'&&+&0&-&\\\hline &&&\dfrac{1}{2}\\y&&\nearrow&&\searrow&\\&0&&&&-\dfrac{1}{2}\\\hline\end{array}
Dựa vào $BBT \Rightarrow \underset{(-\infty;1)}{max \ } y=\dfrac{1}{2} \Leftrightarrow x=-1$
$f)y=-x-\dfrac{1}{x+5}, x <-5\\ y'=-1+\dfrac{1}{(x+5)^2}\\ =\dfrac{1-(x+5)^2}{(x+5)^2}\\ =\dfrac{-x^2 - 10 x - 24}{(x+5)^2}\\ y'=0 \Leftrightarrow x=-4,x=-6\\ BBT:$\begin{array}{|c|ccccccccc|} \hline x&-\infty&&-6&&-5\\\hline y'&&-&0&+&\\\hline &+\infty&&&&+\infty\\y&&\searrow&&\nearrow&\\&&&7\\\hline\end{array}
Dựa vào $BBT \Rightarrow \underset{(-\infty;-5)}{min \ } y=7 \Leftrightarrow x=-5.$
