Đáp án:

$4)min_y=0 \Leftrightarrow x=0$

$7)$Hàm số không có min/max

$8)max_y=4 \Leftrightarrow x=1.$

Giải thích các bước giải:

$4)\\ y=5x^6-24x^5+30x^4 \ \ \ \ D=\mathbb{R}\\ y'=30x^5-120x^4+120x^3\\ =30x^3(x^2-4x+4)\\ =30x^3(x-2)^2\\ y'=0 \Leftrightarrow x=0,x=2(\text{Nghiệm bội chẵn})\\ BBT:$

\begin{array}{|c|ccccccccc|} \hline x&-\infty&&&&0&&2&&\infty\\\hline y'&&&-&&0&+&0&+&\\\hline &+\infty&&&&&&&&+\infty\\&&\searrow&&&&&&\nearrow&\\&&&&&&&32\\y&&&&\searrow&&\nearrow&\\&&&&&0\\\hline\end{array}

Dựa vào $BBT \Rightarrow min_y=0 \Leftrightarrow x=0$

$7)\\ y=(x-3)\sqrt{x^2+1} \ \ \ \ D=\mathbb{R}\\ y'=(x-3)'\sqrt{x^2+1} +(x-3)\sqrt{x^2+1}'\\ =\sqrt{x^2+1} +(x-3).\dfrac{x}{\sqrt{x^2+1}}\\ =\dfrac{x^2+1+x(x-3)}{\sqrt{x^2+1}}\\ =\dfrac{2 x^2 - 3 x + 1}{\sqrt{x^2+1}}\\ y'=0 \Leftrightarrow x=1;x=\dfrac{1}{2}\\ BBT:$

\begin{array}{|c|ccccccccc|} \hline x&-\infty&&\dfrac{1}{2}&&1&&\infty\\\hline y'&&+&0&-&0&+&\\\hline &&&\dfrac{-5\sqrt{5}}{4}&&&&+\infty\\y&&\nearrow&&\searrow&&\nearrow\\&-\infty&&&&-2\sqrt{2}\\\hline\end{array}

Dựa vào $BBT \Rightarrow $Hàm số không có min/max

$8)\\ y=\dfrac{2x+2}{\sqrt{x^2-x+1}} \ \ \ \ D=\mathbb{R}\\ y'=\dfrac{(2x+2)'\sqrt{x^2-x+1}-(2x+2)\sqrt{x^2-x+1}'}{x^2-x+1}\\ =\dfrac{2\sqrt{x^2-x+1}-\dfrac{(2x+2)(2x-1)}{2\sqrt{x^2-x+1}}}{x^2-x+1}\\ =\dfrac{4(x^2-x+1)-(2x+2)(2x-1)}{2(x^2-x+1)^\tfrac{3}{2}}\\ =\dfrac{6 - 6 x}{2(x^2-x+1)^\tfrac{3}{2}}\\ y'=0 \Leftrightarrow x=1\\ BBT:$

\begin{array}{|c|ccccccccc|} \hline x&-\infty&&1&&\infty\\\hline y'&&+&0&-&\\\hline &&&4\\y&&\nearrow&&\searrow&\\&-2&&&&2\\\hline\end{array}

Dựa vào $BBT \Rightarrow max_y=4 \Leftrightarrow x=1.$