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Функције – знак функције 3


Задаци


Текст задатака објашњених у видео лекцији:

пр.9) $y = \frac{{1 - \ln x}}{{1 + \ln x}}$


пр.9)

\[\begin{gathered}
Df:1 + \ln x \ne 0 \cap x > 0 \hfill \\
\ln x \ne - 1 \hfill \\
{e^{ - 1}} \ne x \hfill \\
x \ne \frac{1}{e} \hfill \\
Df:x \in \left( {0;\frac{1}{e}} \right) \cup \left( {\frac{1}{e}; + \infty } \right) \hfill \\
\end{gathered} \]

нуле:

\[\begin{gathered}
y = 0 \hfill \\
\frac{{1 - \ln x}}{{1 + \ln x}} = 0 \hfill \\
1 - \ln x = 0 \hfill \\
\ln x = 1 \hfill \\
{\log _e}x = 1 \hfill \\
{e^1} = x \hfill \\
x = e \in Df \hfill \\
\end{gathered} \]

знак:

\[\begin{gathered}
\ln x = {\log _e}x \hfill \\
e > 1 \hfill \\
\end{gathered} \]

429 png

\[\begin{gathered}
{\log _a}a = 1 \hfill \\
\ln e = 1 \hfill \\
\end{gathered} \]

\[\begin{gathered}
1 - \ln > 0 \hfill \\
\ln x < 1 \hfill \\
\ln x < \ln e \hfill \\
x < e \hfill \\
\hfill \\
1 - \ln < 0 \hfill \\
\ln x > 1 \hfill \\
\ln x > \ln e \hfill \\
x > e \hfill \\
\hfill \\
1 + \ln x > 0 \hfill \\
\ln x > - 1 \hfill \\
\ln x > - \ln e \hfill \\
\ln x > \ln {e^{ - 1}} \hfill \\
x > \frac{1}{e} \hfill \\
\hfill \\
1 + \ln x < 0 \hfill \\
x < \frac{1}{e} \hfill \\
\end{gathered} \]

 434 png

$y > 0$ за $x \in \left( {\frac{1}{e},e} \right)$

$y < 0$ за $x \in \left( {0;\frac{1}{e}} \right) \cup \left( {e; + \infty } \right)$

 

 

 

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