\problemdata     
{S01-PS6-4}           
{Functions: inj/sur/bij}              
{F97, F98-PS7, S01-PS6} 		
{S01 Karger}

\begin{problem}


For each of the functions below, determine whether the
function is injective, whether it is surjective, and whether it is
bijective.  No proofs are required.

\begin{problemparts}
\problempart

 The function $f : \mathbb{R} \mapsto
\mathbb{R}$ defined by $f(x)=e^x$

\solution{ The function is injective because $e^x = e^y$ if and only
if $x = y$.  The function is not surjective, because $e^x$ never takes
on non-positive values.  The function is not bijective, because it is
not surjective.  }

\problempart

The function $f : \mathbb{R} \mapsto
\mathbb{R}^+$ defined by $f(x)=e^x$.  (Here $\mathbb{R}^+$ denotes the
set of positive real numbers.)

\solution{The function is injective as before, but is now also surjective
because for every $y \in \mathbb{R}^+$, there exists an $x$ (namely $x
= \ln y$) such that $y = e^x$.  Therefore, the function is bijective.}

\problempart


The function $f : \mathbb{R} \mapsto
\mathbb{R}$ defined by $(x+1)/(x^2+1)$

\solution{ The function is not injective, because $f(0) = f(1) = 1$.
The function is not surjective, because the function never takes on
large positive or negative values.  The function is not bijective
because it is neither injective nor surjective.}

\problempart

The function $f : \mathbb{R} \mapsto
\mathbb{R}$ defined by

\begin{eqnarray*}
f(x) & = & \left\{
\begin{array}{cl}
1/x & \text{for $x \neq 0$} \\
0 & \text{for $x = 0$.}
\end{array}
\right.
\end{eqnarray*}

\solution{ The function is injective, because $f(x) = f(y)$ if and
only if $x = y$.  The function is surjective, because for all $y \in
\mathbb{R}$, there exists an $x$ (namely $x = 1 / y$ or $x = 0$) such
that $f(x) = y$.  Therefore, the function is also bijective.}

\end{problemparts}

\end{problem}