Функције – примена извода 2

Текст задатака објашњених у видео лекцији:

Пр.3)   Одредити једначину тангенте и нормале криве

            $y = \frac{{2x + 1}}{{x + 1}}$ у тачки пресека криве са $y$-осом.

Пр.4)   Одредити угао под којим се секу параболе $y = {\left( {x - 2} \right)^2}$

            и $y =  - 4 + 6x - {x^2}$.

Пр.3) 

\[\begin{gathered}
y = \frac{{2x + 1}}{{x + 1}} \hfill \\
x = 0 \hfill \\
y = \frac{{2 \cdot 0 + 1}}{{0 + 1}} = 1 \hfill \\
T\left( {{x_0};{y_0}} \right) = T\left( {0;1} \right) \hfill \\
f'\left( x \right) = \frac{{{{\left( {2x + 1} \right)}^\prime }\left( {x + 1} \right) - {{\left( {x + 1} \right)}^\prime }\left( {2x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}} = \frac{{2\left( {x + 1} \right) - \left( {2x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}} = \hfill \\
= \frac{{2x + 2 - 2x - 1}}{{{{\left( {x + 1} \right)}^2}}} = \frac{1}{{{{\left( {x + 1} \right)}^2}}} \hfill \\
f'\left( {{x_0}} \right) = \frac{1}{{{{\left( {{x_0} + 1} \right)}^2}}} \hfill \\
f'\left( 0 \right) = \frac{1}{{{{\left( {0 + 1} \right)}^2}}} = 1 \hfill \\
\hfill \\
t:y - 1 = 1\left( {x - 0} \right) \hfill \\
t:y = x + 1 \hfill \\
\hfill \\
n:y - {y_0} = - \frac{1}{{f'\left( {{x_0}} \right)}}\left( {x - {x_0}} \right) \hfill \\
n:y - 1 = - \frac{1}{1}\left( {x - 0} \right) \hfill \\
n:y - 1 = - x + 1 \hfill \\
\end{gathered} \]

Пр.4)  

\[\begin{gathered}
y = {\left( {x - 2} \right)^2} \hfill \\
\underline {y = - 4 + 6x - {x^2}} \hfill \\
{\left( {x - 2} \right)^2} = - 4 + 6x - {x^2} \hfill \\
{\left( {x - 2} \right)^2} + 4 - 6x + {x^2} = 0 \hfill \\
{x^2} - 4x + 4 + 4 - 6x + {x^2} = 0 \hfill \\
2{x^2} - 10x + 8 = 0|:2 \hfill \\
{x^2} - 5x + 4 = 0 \hfill \\
{x_{1/2}} = \frac{{5 \pm \sqrt {25 - 16} }}{2} \hfill \\
{x_{1/2}} = \frac{{5 \pm 3}}{2} \hfill \\
\begin{array}{*{20}{c}}
{{x_1} = 1}&{}&{}&{{x_2} = 4} \\
{{y_1} = {{\left( {1 - 2} \right)}^2}}&{}&{}&{{y_2} = {{\left( {4 - 2} \right)}^2}} \\
{{y_1} = 1}&{}&{}&{{y_2} = 4} \\
{A\left( {1;1} \right)}&{}&{}&{B\left( {4;4} \right)}
\end{array} \hfill \\
\hfill \\
t:y - {y_0} = f'\left( {{x_0}} \right)\left( {x - {x_0}} \right)\begin{array}{*{20}{c}}
{}&{}&{}
\end{array}A\left( {1;1} \right) \hfill \\
\hfill \\
\begin{array}{*{20}{c}}
\begin{gathered}
{p_1}:f\left( x \right) = y = {\left( {x - 2} \right)^2} \hfill \\
f'\left( x \right) = 2\left( {x - 2} \right){\left( {x - 2} \right)^\prime } = 2x - 4 \hfill \\
f'\left( {{x_0}} \right) = 2 \cdot 1 - 4 = - 2 \hfill \\
\end{gathered} &{}&{}&{}&\begin{gathered}
{p_2}:f\left( x \right) = y = - 4 + 6x - {x^2} \hfill \\
f'\left( x \right) = 6 - 2x \hfill \\
f'\left( {{x_0}} \right) = 6 - 2 \cdot 1 = 4 \hfill \\
\end{gathered} \\
\begin{gathered}
{t_1}:y - 1 = - 2\left( {x - 1} \right) \hfill \\
y = - 2x + 2 + 1 \hfill \\
y = - 2x + 3 \hfill \\
{k_{{t_1}}} = - 2 \hfill \\
\end{gathered} &{}&{}&{}&\begin{gathered}
{t_2}:y - 1 = 4\left( {x - 1} \right) \hfill \\
y = 4x - 4 + 1 \hfill \\
y = 4x - 3 \hfill \\
{k_{{t_2}}} = 4 \hfill \\
\end{gathered}
\end{array} \hfill \\
\hfill \\
tg\varphi = \left| {\frac{{{k_1} - {k_2}}}{{1 + {k_1}{k_2}}}} \right| = \left| {\frac{{ - 2 - 4}}{{1 - 2 \cdot 4}}} \right| = \frac{6}{7} \hfill \\
\end{gathered} \]

\[\varphi  = arctg\frac{6}{7}\]