Функције – примена извода 1

Текст задатака објашњених у видео лекцији:

Пр.1)   Одредити једначину тангенте и нормале криве

            $f\left( x \right) = 2{x^2} - 3x + 7$ у тачки $A\left( {2, - 3} \right)$.

Пр.2)   Одредити једначину тангенте графика функције

            $y =  - {x^2} - 2$ која је паралелна прави $p:4x - y + 1 = 0$.

Пр.1)

\[\begin{gathered}
f\left( x \right) = 2{x^2} - 3x + 7;A\left( {2, - 3} \right) \hfill \\
\hfill \\
t:y - {y_0} = f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) \hfill \\
f'\left( x \right) = 4x - 3 \hfill \\
f'\left( 2 \right) = 4 \cdot 2 - 3 \hfill \\
f'\left( 2 \right) = 5 \hfill \\
t:y + 3 = 5\left( {x - 2} \right) \hfill \\
t:y + 3 = 5x - 10 \hfill \\
t:y = 5x - 13 \hfill \\
\hfill \\
n:y - {y_0} = - \frac{1}{{f'\left( {{x_0}} \right)}}\left( {x - {x_0}} \right) \hfill \\
n:y + 3 = - \frac{1}{5}\left( {x - 2} \right) \hfill \\
n:y = - \frac{1}{5}\left( {x - 2} \right) - 3 \hfill \\
n:y = - \frac{1}{5}x + \frac{2}{5} - 3 \hfill \\
n:y = - \frac{1}{5}x - \frac{{13}}{5} \hfill \\
\end{gathered} \]

Пр.2)

\[\begin{gathered}
y = - {x^2} - 2 \hfill \\
p:4x - y + 1 = 0 \hfill \\
\hfill \\
t\parallel p \leftrightarrow {k_t} = {k_p} \hfill \\
p:4x - y + 1 = 0 \hfill \\
p:y = 4x + 1 \Rightarrow {k_p} = 4 \Rightarrow {k_t} = 4 \hfill \\
\hfill \\
t:y - {y_0} = f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) \hfill \\
\hfill \\
f\left( x \right) = - {x^2} - 2 \hfill \\
f'\left( x \right) = - 2x \hfill \\
\hfill \\
f'\left( {{x_0}} \right) = {k_t} = 4 \hfill \\
f'\left( {{x_0}} \right) = - 2{x_0} = 4 \Rightarrow {x_0} = - 2 \hfill \\
\hfill \\
{y_0} = - {x_0}^2 - 2 \hfill \\
{y_0} = - {\left( { - 2} \right)^2} - 2 \hfill \\
{y_0} = - 4 - 2 \hfill \\
{y_0} = - 6 \hfill \\
A\left( { - 2; - 6} \right) \hfill \\
\hfill \\
t:y - {y_0} = f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) \hfill \\
t:y + 6 = 4\left( {x + 2} \right) \hfill \\
t:y + 6 = 4x + 8 \hfill \\
t:y = 4x + 2 \hfill \\
\end{gathered} \]