Функције – пример 5

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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) \[\begin{gathered}
  {g^{ - 1}}\left( x \right) =  \hfill \\
  {g^{ - 1}}\left( {g\left( x \right)} \right) = x \hfill \\
  {g^{ - 1}}\left( {\frac{{x - 5}}{2}} \right) = x \hfill \\
  \frac{{x - 5}}{2} = t \hfill \\
  x - 5 = 2t \hfill \\
  x = 2t + 5 \hfill \\
   \hfill \\
  {g^{ - 1}}\left( t \right) = 2t + 5 \hfill \\
  {g^{ - 1}}\left( x \right) = 2x + 5 \hfill \\
\end{gathered} \]

\[\begin{gathered}
  {f^{ - 1}} \circ g\left( x \right) = {f^{ - 1}}\left( {g\left( x \right)} \right) = {f^{ - 1}}\left( {\frac{{x - 5}}{2}} \right) = \frac{{\frac{{x - 5}}{2} - 3}}{5} = \frac{{\frac{{x - 5 - 6}}{2}}}{5} =  \hfill \\
   = \frac{{x - 11}}{{10}} \hfill \\
  {g^{ - 1}} \circ {f^{ - 1}}\left( x \right) = {g^{ - 1}}\left( {{f^{ - 1}}\left( x \right)} \right) = {g^{ - 1}}\left( {\frac{{x - 3}}{5}} \right) = 2\frac{{x - 3}}{5} + 5 =  \hfill \\
   = \frac{{2x - 6}}{5} + 5 = \frac{{2x - 6 + 25}}{5} = \frac{{2x + 19}}{5} \hfill \\
\end{gathered} \]