Đáp án:
d. \(y' < 0\forall x \ne 3\)
e. \(y' \le 0\forall x \in R\)
Giải thích các bước giải:
\(\begin{array}{l}
a.y' = 3{x^2} - 6x\\
y' = 0\\
\to 3{x^2} - 6x = 0\\
\to 3x\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.
\end{array}\)
BXD
x -∞ 0 2 +∞
y' + 0 - 0 +
\(\begin{array}{l}
b.y' = - 4{x^3} + 4x\\
y' = 0 \to - 4{x^3} + 4x = 0\\
\to 4x\left( { - {x^2} + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = - 1\\
x = 1
\end{array} \right.
\end{array}\)
BXD
x -∞ -1 0 1 +∞
y' + 0 - 0 + 0 -
\(\begin{array}{l}
c.DK:{x^2} + 2x > 0\\
\to \left[ \begin{array}{l}
x > 0\\
x < - 2
\end{array} \right.\\
y' = \dfrac{{2x + 2}}{{2\sqrt {{x^2} + 2x} }} = \dfrac{{x + 1}}{{\sqrt {{x^2} + 2x} }}\\
y' = 0 \to \dfrac{{x + 1}}{{\sqrt {{x^2} + 2x} }} = 0\\
\to x = - 1\left( l \right)
\end{array}\)
Do x=-1 không thuộc TXĐ
BXD
x -∞ -2 0 +∞
y' - // \\\\\\\\\\ // +
\(\begin{array}{l}
d.DK:x \ne 3\\
y' = \dfrac{{2\left( {x - 3} \right) - 2x - 1}}{{{{\left( {x - 3} \right)}^2}}}\\
= \dfrac{{ - 7}}{{{{\left( {x - 3} \right)}^2}}}\\
Do:\left\{ \begin{array}{l}
- 7 < 0\\
{\left( {x - 3} \right)^2} > 0\forall x \ne 3
\end{array} \right.\\
\to y' < 0\forall x \ne 3
\end{array}\)
\(\begin{array}{l}
e.y' = - {x^2} + 2x - 1\\
= - {\left( {x - 1} \right)^2}\\
Do:{\left( {x - 1} \right)^2} \ge 0\forall x \in R\\
\to - {\left( {x - 1} \right)^2} \le 0\forall x \in R\\
\to y' \le 0
\end{array}\)
