Đáp án:
Giải thích các bước giải:
a) `y=\frac{5x^2}{2}-x-0,25`
TXĐ: `D=\mathbb{R}`
`y'=5x-1`
`y'=0 \Rightarrow x =1/5`
Ta có bảng sau:
\(\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}\dfrac{1}{5}&\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}-&\text{0}&\text{}+&\text{}\\\hline \text{$y$}&\text{}+\infty&\text{}&\text{}&\text{}&\text{}+\infty\\&\text{}&\text{$\searrow$}&\text{}&\text{}\nearrow&\text{}\\&\text{}&\text{}&\text{}-\dfrac{7}{20}&\text{}&\\\hline \end{array}\)
Vậy HS đồng biến tại `(1/5;+\infty)` và nghịch biến tại `(-\infty;1/5)`
c) `y=\frac{x^3}{3}-\frac{x^2}{2}-6x+\sqrt{2}`
TXĐ: `D=\mathbb{R}`
`y'=x^2-x-6`
`y'=0 ⇒` \(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
Ta có bảng biến thiên:
\(\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}-2&\text{}&\text{}3&\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}+&\text{0}&\text{}-&\text{}0&\text{}+&\\\hline \text{$y$}&\text{}&\text{}&\text{}\dfrac{22+3\sqrt{2}}{3}&\text{}&\text{}&\text{}&+\infty\\&\text{}&\text{$\nearrow$}&\text{}&\text{}\searrow&\text{}&\text{}\nearrow\\&\text{$-\infty$}&\text{}&\text{}&\text{}&\dfrac{-27+2\sqrt{2}}{2}\\\hline \end{array}\)
Vậy HS đồng biến tại `(-\infty;-2)` và `(3;+\infty)`
Nghịch biến tại `(-2;3)`
e) `y=\frac{x^4}{4}-8x^2+3/4`
TXĐ: `D=\mathbb{R}`
`y'=x^3-16x`
`y'=0⇒` \(\left[ \begin{array}{l}x=-4\\x=0\\x=4\end{array} \right.\)
Ta có BBT:
\(\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}-4&\text{}&\text{}0&\text{}&4&\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}-&\text{0}&\text{}+&\text{}0&\text{}-&\text{}0&\text{}+&\\\hline \text{$y$}&\text{$+\infty$}&\text{}&\text{}&\text{}&\text{}\dfrac{3}{4}&\text{}&\text{}&&+\infty\\&\text{}&\text{$\searrow$}&\text{}&\text{}\nearrow&\text{}&\text{}\searrow&\text{}&\nearrow\\&\text{}&\text{}&\text{}-\dfrac{253}{4}&\text{}&&\text{}&\text{}-\dfrac{253}{4}&&\\\hline \end{array}\)
Vậy HS đồng biến tại `(-4;0)` và `(4;+\infty)`
Nghịch biến tại `(-\infty;-4)` và `(0;4)`
