Đáp án:
a. \(y = - 3x + 3\)
Giải thích các bước giải:
\(\begin{array}{l}
y' = 3{x^2} - 6x\\
a.y'\left( {{x_0}} \right) = k = 3{x_0}^2 - 6{x_0}\\
Thay:{x_0} = 1\\
\to k = 3 - 6 = - 3\\
\to PTTT:y = - 3\left( {x - 1} \right) + 0\\
\to y = - 3x + 3\\
b.Do:{x_0} = 2\\
\to {y_0} = - 2\\
Thay:{x_0} = 2\\
\to k = {3.2^2} - 6.2 = 3.4 - 12 = 0\\
\to PTTT:y = 0\left( {x - 2} \right) - 2\\
\to y = - 2\\
c.Do:{y_0} = 2\\
\to 2 = {x_0}^3 - 3{x_0}^2 + 2\\
\to \left[ \begin{array}{l}
{x_0} = 0\\
{x_0} = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
k = {3.0^2} - 6.0 = 0\\
k = {3.3^2} - 6.3 = 9
\end{array} \right.\\
\to PTTT:\left[ \begin{array}{l}
y = 0.x + 2\\
y = 9\left( {x - 3} \right) + 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = 2\\
y = 9x - 25
\end{array} \right.\\
d.Do:k = 3\\
\to 3{x_0}^2 - 6{x_0} = 3\\
\to \left[ \begin{array}{l}
{x_0} = 1 + \sqrt 2 \\
{x_0} = 1 - \sqrt 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
{y_0} = - \sqrt 2 \\
{y_0} = \sqrt 2
\end{array} \right.\\
\to PTTT:\left[ \begin{array}{l}
y = 3\left( {x - 1 - \sqrt 2 } \right) - \sqrt 2 \\
y = 3\left( {x - 1 + \sqrt 2 } \right) + \sqrt 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = 3x - 3 - 4\sqrt 2 \\
y = 3x - 3 + 4\sqrt 2
\end{array} \right.
\end{array}\)
