Đáp án:
Giải thích các bước giải:
`y=(x^2-3)^2`
TXĐ: `D=\mathbb{R}`
`y'=4x^3-12x`
`y'=0⇒` \(\left[ \begin{array}{l}x=-\sqrt{3}\\x=\sqrt{3}\\x=0\end{array} \right.\)
Ta có bảng sau:
\(\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}-\sqrt{3}&\text{}&\text{}0&\text{}&\text{}\sqrt{3}&\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}-&\text{0}&\text{}+&\text{0}&\text{}-&\text{}0&\text{}+&\\\hline \text{$y$}&\text{}+\infty&\text{}&\text{}&\text{}&\text{}9&\text{}&\text{}&\text{}&\text{$+\infty$}\\&\text{}&\text{$\searrow$}&\text{}&\text{}\nearrow&\text{}&\text{}\searrow&\text{}&\nearrow\\&\text{$$}&\text{}&\text{}0&\text{}&\text{}&\text{}&\text{}0&\text{}&\\\hline \end{array}\)
Vậy HS đạt cực đại tại `x=0,y_{CĐ}=9`
