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\problemgrouptitle{Inverse Functions}

\begin{problemblock}[2]{Show that the given function is one-to-one and find its inverse.}

\item $f(x)=2x^3+5$   
\answer{
$\displaystyle f^{-1}(x)=\sqrt [3]{\frac{x-5}{2}}$
}

\item $\displaystyle f(x)=\frac{2x}{3x-1}$   
\answer{
$\displaystyle f^{-1}(x)=\frac{x}{3x-2}$
}

\item $f(x)=\sqrt{2x-3}$   
\answer{
$f^{-1}(x)=\frac{1}{2} \left(x^2+3\right)$
}

\item $f(x)={\left(x^3+2\right)^7}$   
\answer{
$f^{-1}(x)=\sqrt[3]{x^{1/7}-2}$
}

\end{problemblock}

\begin{problemblock}[2]{In the following problems
    \begin{enumerate}
    \item Show that $f$ is one-to-one.
    \item Find $f^{-1}(x)$ and state the domain and range of $f^{-1}$.
    \end{enumerate}
  }
  
\item $f(x)=x^5+4$   
\answer{
    (b) $f^{-1}(x)=\sqrt[5]{x-4}$,
    Domain=$\left(-\infty,\infty\right)$,
    Range=$\left(-\infty,\infty\right)$
}

\item $\displaystyle f(x)=\frac{x+3}{2x+1}$   
\answer{
  (b) $\displaystyle f^{-1}(x)=\frac{3-x}{2x-1}$, 
  Domain=$\mathset{R}-\left\{\frac{1}{2}\right\}$,
  Range=$\mathset{R}-\left\{-\frac{1}{2}\right\}$
}

\item $\displaystyle f(x)=\frac{\sqrt{x}+1}{\sqrt{x}+2}$   
\answer{
  (b) $\displaystyle f^{-1}(x)=\left(\frac{1-2x}{x-1}\right)^2$, 
  Domain=$\left[\frac{1}{2},1\right)$,
  Range=$\left[0,\infty\right)$\\
  Note that for the domain when solving $x=f(y)$ for $y$ you get $\sqrt{x}=\frac{1-2y}{y-1}$. Since $\sqrt{x}\ge 0$ this implies $\frac{1-2y}{y-1}\ge0$. Exchanging variables means $\frac{1-2x}{x-1}\ge0$ which restricts the domain of $f^{-1}$.
}
\end{problemblock}

\begin{problemblock}{Find $\left(f^{-1}\right)'(a)$ for the given $a$.}

\item $f(x)=2x^3+5,  a=7$   
\answer{
$\left(f^{-1}\right)'(7)=\frac{1}{6}$
}

\item $f(x)=x^5+\sin{x}+2x+2,  a=2$   
\answer{
$\left(f^{-1}\right)'(2)=\frac{1}{3}$
}


\item $f(x)=\sqrt[3]{x}+x+1,  a=11$   
\answer{
$\left(f^{-1}\right)'(11)=\frac{12}{13}$
}
\end{problemblock}

\item Suppose $f$ and $g$ are invertible functions.  Consider the composite function $h=f\circ g$ defined by $h(x)=f\left(g(x)\right)$\,.
\begin{enumerate}
\item Show that $h$ is one-to-one.
\item Find $h^{-1}(x)$.
\end{enumerate}
\answer{
  \begin{enumerate}
  \item $h(x_1)=h(x_2)\implies f\left(g(x_1)\right)=f\left(g(x_2)\right)\implies f^{-1}\left[ f\left(g(x_1)\right)\right]=f^{-1}\left[ f\left(g(x_2)\right)\right]\implies~g(x_1)=g(x_2)\implies x_1=x_2$
  \item $h^{-1}(x)=g^{-1}\left(f^{-1}(x)\right)$ so $h^{-1}=g^{-1}\circ f^{-1}$
  \end{enumerate}
}

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