%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                                           %
%  Function Algebra                         %
%                                           %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problemgrouptitle{Function Algebra}

\item Suppose that $f(x)=\displaystyle\frac{1}{x+2}$.  Determine:
\begin{multicols}{2}
\begin{enumerate}
\item $f(x+2)$	\columnbreak
\item $f(f(x))$
\end{enumerate}
\end{multicols}
\answer{ 
\begin{inparaenum}[(a)]
\item $f(x+2)=\displaystyle\frac{1}{x+4}$
\item $f(f(x))=\displaystyle\frac{x+2}{2x+5}$
\end{inparaenum}
}

\begin{problemblock}{Find the composite functions $f\circ g$ and $g\circ f$ and their domains for the given functions.}

\item $\displaystyle f(x)=2x^3$, \  $\displaystyle g(x)=\sqrt{x^2+3}$.
\answer{$f\circ g(x)=f(g(x))=2\left(x^2+3\right)^\frac{3}{2}$ with  $D=\mathset{R}$, $g\circ f(x)=g(f(x))=\displaystyle\sqrt{4x^6+3}$ with $D=\mathset{R}$}

\item $\displaystyle f(x)=3x^2+6x+4$, \ $\displaystyle g(x)=3x-2$
\answer{$f\circ g(x)=f(g(x))=3(3x-2)^2+6(3x-2)+4=27x^2-18x+4$ with  $D=\mathset{R}$, $g\circ f(x)=g(f(x))=3(3x^2+6x+4)-2=9x^2+18x+10$ with $D=\mathset{R}$}

\item $\displaystyle f(z)=\sqrt{z^2+5}$, \ $\displaystyle g(z)=\frac{z}{z+1}$
\answer{$\displaystyle f\circ g(z)=f(g(z))=\sqrt{\left(\frac{z}{z+1}\right)^2+5}=\sqrt{\frac{6z^2+10z+5}{(z+1)^2}}$ with  $D=\mathset{R}-\{-1\}$, $\displaystyle g\circ f(z)=g(f(z))=\frac{\sqrt{z^2+5}}{\sqrt{z^2+5}+1}$ with $D=\mathset{R}$}

\item $\displaystyle f(x)=\frac{2x+5}{x-4}$, \ $\displaystyle g(x)=x^2+3$
\answer{$\displaystyle f\circ g(x)=f(g(x))=\frac{2\left(x^2+3\right)+5}{\left(x^2+3\right)-4}=\frac{2x^2+11}{x^2-1}$ with  $D=\mathset{R}-\{-1,1\}$, $\displaystyle g\circ f(x)=g(f(x))=\left(\frac{2x+5}{x-4}\right)^2+3=\frac{7x^2-4x+73}{(x-4)^2}$ with $D=\mathset{R}-\{4\}$}
\end{problemblock}

\item Find two functions $f(x)$ and $g(x)$ such that $f(g(x))=\sqrt{x^2+1}-3$.
\answer{$f(x)=\sqrt{x}-3$, $g(x)=x^2+1$.}

\item Use function composition to solve the equation $\displaystyle \left(x^2-5\right)^2+7x^2-23=0$.  (Hint: Note the equation can be rewritten as $\displaystyle \left(x^2-5\right)^2+7\left(x^2-5\right)+12=0$.)
\answer{$\displaystyle\left\{\pm1,\pm\sqrt{2}\right\}$}

%%% Local Variables:
%%% mode: latex
%%% TeX-master: "../master"
%%% End:
