Đáp án:
16A
Giải thích các bước giải:
`y=-x^4+2x^2+5/4`
TXĐ: `D=\mathbb{R}`
`y'=-4x^3+4x`
`y'=0⇒` \(\left[ \begin{array}{l}x=1\\x=-1\\x=0\end{array} \right.\)
Ta có bảng sau:
\(\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}-1&\text{}&\text{}0&\text{}&\text{}1&\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}+&\text{0}&\text{}-&\text{0}&\text{+}&\text{}0&\text{}-&\\\hline \text{$y$}&\text{}&\text{}&\text{}\dfrac{9}{4}&\text{}&\text{}&\text{}&\text{$\dfrac{9}{4}$}\\&\text{}&\text{$\nearrow$}&\text{}&\text{}\searrow&\text{}&\text{}\nearrow&\text{}&\searrow\\&\text{$-\infty$}&\text{}&\text{}&\text{}&\text{}\dfrac{5}{4}&\text{}&\text{}&\text{}&-\infty\\\hline \end{array}\)
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