Функционална једначина и инверзна функција.
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Текст задатака објашњених у видео лекцији.
Пр.5 $f\left( x \right) = 5x + 3$ и $g\left( {2x - 1} \right) = x - 3$. Одредити ${f^{ - 1}} \circ g\left( x \right) = $ и ${g^{ - 1}} \circ {f^{ - 1}}\left( x \right) = $.
Пр.5
1) Израчунамо ${f^{ - 1}}\left( x \right)$. Знамо да је ${f^{ - 1}}\left( {f\left( x \right)} \right) = x$ онда ${f^{ - 1}}\left( {f\left( {5x + 3} \right)} \right) = x$.
Замена:
\[\begin{gathered}
5x + 3 = t \hfill \\
5x = t - 3 \hfill \\
x = \frac{{t - 3}}{5} \hfill \\
\end{gathered} \]
\[\begin{gathered}
{f^{ - 1}}\left( t \right) = \frac{{t - 3}}{5} \hfill \\
{f^{ - 1}}\left( x \right) = \frac{{x - 3}}{5} \hfill \\
\end{gathered} \]
2) \[\begin{gathered}
g\left( x \right) = \hfill \\
g\left( {2x - 1} \right) = x - 3 \hfill \\
2x - 1 = t \hfill \\
2x = t + 1 \hfill \\
x = \frac{{t + 1}}{2} \hfill \\
\hfill \\
g\left( t \right) = \frac{{t + 1}}{2} - 3 \hfill \\
g\left( t \right) = \frac{{t + 1 - 6}}{2} \hfill \\
g\left( t \right) = \frac{{t - 5}}{2} \hfill \\
g\left( x \right) = \frac{{x - 5}}{2} \hfill \\
\end{gathered} \]
3) \[\begin{gathered}
{g^{ - 1}}\left( x \right) = \hfill \\
{g^{ - 1}}\left( {g\left( x \right)} \right) = x \hfill \\
{g^{ - 1}}\left( {\frac{{x - 5}}{2}} \right) = x \hfill \\
\frac{{x - 5}}{2} = t \hfill \\
x - 5 = 2t \hfill \\
x = 2t + 5 \hfill \\
\hfill \\
{g^{ - 1}}\left( t \right) = 2t + 5 \hfill \\
{g^{ - 1}}\left( x \right) = 2x + 5 \hfill \\
\end{gathered} \]
\[\begin{gathered}
{f^{ - 1}} \circ g\left( x \right) = {f^{ - 1}}\left( {g\left( x \right)} \right) = {f^{ - 1}}\left( {\frac{{x - 5}}{2}} \right) = \frac{{\frac{{x - 5}}{2} - 3}}{5} = \frac{{\frac{{x - 5 - 6}}{2}}}{5} = \hfill \\
= \frac{{x - 11}}{{10}} \hfill \\
{g^{ - 1}} \circ {f^{ - 1}}\left( x \right) = {g^{ - 1}}\left( {{f^{ - 1}}\left( x \right)} \right) = {g^{ - 1}}\left( {\frac{{x - 3}}{5}} \right) = 2\frac{{x - 3}}{5} + 5 = \hfill \\
= \frac{{2x - 6}}{5} + 5 = \frac{{2x - 6 + 25}}{5} = \frac{{2x + 19}}{5} \hfill \\
\end{gathered} \]