Đáp án:
$1. min_y=-4 \Leftrightarrow x=-1\\ 2. max_y=1 \Leftrightarrow x=1\\ 3. max_y=10; min_y=1\\ 4. max_y=-5 \Leftrightarrow x=-1\\ 5. min_y=8 \Leftrightarrow x=4\\ 6. max_y=\dfrac{1}{3}; min_y=2\\ 7. max_y=1; min_y=-5+2\sqrt{5}$
Giải thích các bước giải:
$1. y=x^2+2x-3\\ =x^2+2x+1-4\\ =(x+1)^2-4 \ge -4 \ \forall \ x\\ \Rightarrow min_y=-4$
Dấu "=" xảy ra $\Leftrightarrow x=-1$
$2. y=4x^3-3x^4\\ y'=12x^2-12x^3=12x^2(1-x)\\ y'=0 \Leftrightarrow x=0;x=1\\ BBT:\\ \begin{array}{|c|ccc|} \hline x&-\infty&&0&&1&&\infty\\\hline y'&&+&0&+&0&-&\\\hline &&&&&1&&\\y&&&\nearrow&&&\searrow\\&-\infty&&&&&&-\infty\\\hline\end{array}$
Dựa vào bảng $\Rightarrow max_y=1$ tại $x=1$
$3. y=x^4-2x^2+2\\ y'=4x^3-4x=4x(x^2-1)=4x(x-1)(x+1)\\ y'=0 \Leftrightarrow x=0;x=\pm 1\\ y(-2)=10\\ y(-1)=1\\ y(0)=2\\ y(1)=1\\ \Rightarrow \underset{x\in[-2;1]}{max_y}=y(-2)=10; \underset{x\in[-2;1]}{min_y}=y(-1)=y(1)=1\\ 4. y=\dfrac{x^2-3x+1}{x}\\ y'=\dfrac{(2x-3)x-x^2+3x-1}{x^2}=\dfrac{x^2-1}{x^2}=\dfrac{(x-1)(x+1)}{x^2}\\ y'=0 \Leftrightarrow x=\pm 1\\ \lim_{x \to 0^-} y=-\infty\\ \lim_{x \to -\infty} y=-\infty\\ BBT:\\ \begin{array}{|c|ccccccccc|} \hline x&-\infty&&-1&&0\\\hline y'&&+&0&-&\\\hline &&&-5&&\\y&&\nearrow&&\searrow\\&-\infty&&&&-\infty\\\hline\end{array}$
Dựa vào bảng, $max_y=-5$ tại $x=-1$
$5. \dfrac{x^2+16}{x}\\ y'=\dfrac{2x.x-x^2-16}{x^2}=\dfrac{x^2-16}{x^2}=\dfrac{(x-4)(x+4)}{x^2}\\ y'=0 \Leftrightarrow x=\pm 4\\ \lim_{x \to 0^+} y=+\infty\\ \lim_{x \to +\infty} y=+\infty\\ BBT:\\ \begin{array}{|c|ccccccccc|} \hline x&0&&4&&\infty\\\hline y'&&-&0&+&\\\hline &+\infty&&&&+\infty\\y&&\searrow&&\nearrow\\&&&8&&\\\hline\end{array}$
Dựa vào bảng, $min_y=8$ tại $x=4$
$6. y=\dfrac{x+1}{x-1}\\ y'=\dfrac{x-1-x-1}{(x-1)^2}=\dfrac{-2}{(x-1)^2}<0 \ \forall x \in D\\ \Rightarrow \underset{x\in[-2;3]}{max_y}=y(-2)=\dfrac{1}{3}; \underset{x\in[-2;3]}{min_y}=y(3)=2\\ 7.y= \dfrac{x^2-3x+1}{x+1}\\ y'= \dfrac{(2x-3)(x+1)-x^2+3x-1}{x+1}= \dfrac{x^2+2x-4}{x+1}= \dfrac{(x+1-\sqrt{5})(x+1+\sqrt{5})}{x+1}\\ y'=0 \Leftrightarrow x=-1 \pm \sqrt{5}\\ y(1)=-\dfrac{1}{2}\\ y(-1+ \sqrt{5})=-5+2\sqrt{5}\\ y(4)=1\\ \Rightarrow \underset{x\in[1;4]}{max_y}=y(4)=1; \underset{x\in[1;4]}{min_y}=y(-1+ \sqrt{5})=-5+2\sqrt{5}$
