Đáp án:
c. \(\left[ \begin{array}{l}
y = - 2\\
y = - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
y' = 4{x^3} - 4x\\
a.Do:{x_0} = 2\\
\to {y_0} = 7\\
\to k = y'\left( 2 \right) = {4.2^3} - 4.2 = 24\\
\to PTTT:y = 24\left( {x - 2} \right) + 7\\
\to t = 24x - 41\\
b.Do:{y_0} = - 1\\
\to \left[ \begin{array}{l}
{x_0} = 0\\
{x_0} = - \sqrt 2 \\
{x_0} = \sqrt 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
k = f'\left( 0 \right) = 0\\
k = f'\left( {\sqrt 2 } \right) = 4\sqrt 2 \\
k = f'\left( { - \sqrt 2 } \right) = - 4\sqrt 2
\end{array} \right.\\
\to PTTT:\left[ \begin{array}{l}
y = 0\left( {x - 0} \right) - 1\\
y = 4\sqrt 2 \left( {x - \sqrt 2 } \right) - 1\\
y = - 4\sqrt 2 \left( {x + \sqrt 2 } \right) - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = - 1\\
y = 4\sqrt 2 x - 9\\
y = - 4\sqrt 2 x - 9
\end{array} \right.\\
c.k = 0\\
\to y'\left( {{x_0}} \right) = 4{x_0}^3 - 4{x_0} = 0\\
\to \left[ \begin{array}{l}
{x_0} = 1\\
{x_0} = - 1\\
{x_0} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
{y_0} = - 2\\
{y_0} = - 2\\
{y_0} = - 1
\end{array} \right.\\
\to PTTT:\left[ \begin{array}{l}
y = 0\left( {x - 1} \right) - 2\\
y = 0\left( {x + 1} \right) - 2\\
y = 0.x - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = - 2\\
y = - 1
\end{array} \right.
\end{array}\)
