Đáp án:
c. \(y = - 3x + 2\)
Giải thích các bước giải:
\(\begin{array}{l}
y' = 3{x^2} - 3\\
a.Do:{x_0} = 3\\
\to {y_0} = {x_0}^3 - 3{x_0} + 2\\
= {3^3} - 3.3 + 2 = 20\\
\to y'\left( {{x_0}} \right) = k = 3{x_0}^2 - 3 = 3.9 - 3 = 24\\
\to PTTT:y = 24\left( {x - 3} \right) + 20\\
\to y = 24x - 52\\
b.Do:{y_0} = 2\\
\to 2 = {x_0}^3 - 3{x_0} + 2\\
\to \left[ \begin{array}{l}
{x_0} = 0\\
{x_0} = \sqrt 3 \\
{x_0} = - \sqrt 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
k = y'\left( 0 \right) = - 3\\
k = y'\left( {\sqrt 3 } \right) = 6\\
k = y'\left( { - \sqrt 3 } \right) = 6
\end{array} \right.\\
\to PTTT:\left[ \begin{array}{l}
y = - 3x + 2\\
y = 6\left( {x - \sqrt 3 } \right) + 2\\
y = 6\left( {x + \sqrt 3 } \right) + 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = - 3x + 2\\
y = 6x - 6\sqrt 3 + 2\\
y = 6x + 6\sqrt 3 + 2
\end{array} \right.\\
c.Do:tt/Ox\\
\to {x_0} = 0\\
\to {y_0} = 2\\
\to k = y'\left( 0 \right) = - 3\\
\to PTTT:y = - 3x + 2
\end{array}\)
