Đáp án:

Giải thích các bước giải:

 b) `y=-x^2+6x-1,7`

TXĐ: `D=\mathbb{R}`

`y'=-2x+6`

`y'=0 \Rightarrow x =3`

Ta có bảng sau:

\(\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}3&\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}+&\text{0}&\text{}-&\text{}\\\hline \text{$y$}&\text{}&\text{}&\text{}\dfrac{73}{10}&\text{}&\text{}\\&\text{}&\text{$\nearrow$}&\text{}&\text{}\searrow&\text{}\\&\text{$-\infty$}&\text{}&\text{}&\text{}&-\infty\\\hline \end{array}\)

Vậy HS đồng biến tại `(-\infty;3)` và nghịch biến tại `(3;+\infty)`

d) `y=4x^3-3x`

TXĐ: `D=\mathbb{R}`

`y'=12x^2-3`

`y'=0 ⇒` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{array} \right.\) 

Ta có bảng biến thiên:

\(\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}-\dfrac{1}{2}&\text{}&\text{}\dfrac{1}{2}&\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}+&\text{0}&\text{}-&\text{}0&\text{}+&\\\hline \text{$y$}&\text{}&\text{}&\text{}1&\text{}&\text{}&\text{}&+\infty\\&\text{}&\text{$\nearrow$}&\text{}&\text{}\searrow&\text{}&\text{}\nearrow\\&\text{$-\infty$}&\text{}&\text{}&\text{}&-1\\\hline \end{array}\)

Vậy HS đồng biến tại `(-\infty;-1/2)` và `(1/2;+\infty)`

Nghịch biến tại `(-1/2;1/2)`

f) `y=x^2-x^4`

TXĐ: `D=\mathbb{R}`

`y'=2x-4x^3`

`y'=0⇒` \(\left[ \begin{array}{l}x=-\dfrac{\sqrt{2}}{2}\\x=0\\x=\dfrac{\sqrt{2}}{2}\end{array} \right.\) 

Ta có BBT:

\(\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}-\dfrac{\sqrt{2}}{2}&\text{}&\text{}0&\text{}&\dfrac{\sqrt{2}}{2}&\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}+&\text{0}&\text{}-&\text{}0&\text{}+&\text{}0&\text{}-&\\\hline \text{$y$}&\text{}&\text{}&\text{}\dfrac{1}{4}&\text{}&\text{}&\text{}&\dfrac{1}{4}\\&\text{}&\text{$\nearrow$}&\text{}&\text{}\searrow&\text{}&\text{}\nearrow&\text{}&\searrow\\&\text{$-\infty$}&\text{}&\text{}&\text{}&0&\text{}&\text{}&&-\infty\\\hline \end{array}\)

Vậy HS đồng biến tại `(-\infty;-\frac{\sqrt{2}}{2})` và `(0;\frac{\sqrt{2}}{2})`

Nghịch biến tại `(-\frac{\sqrt{2}}{2};0)` và `(\frac{\sqrt{2}}{2};+\infty)`