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\problemgrouptitle{Continuity}

\item Define precisely what is meant by the statement ``$f$ is continuous at $x=a$''.
\answer{ ``$f$ is continuous at $x=a$'' if 
\begin{inparaenum}[(a)]
\item $a$ is in the domain of $f$
\item $\displaystyle\lim_{x\to a}f(x)$ exists
\item $\displaystyle\lim_{x\to a}f(x)=f(a)$.
\end{inparaenum}
}

\begin{problemblock}{Use the continuity definition to determine if the function is continuous at the given value.  If the function is discontinuous there, decide whether it is a removable, jump, or infinite discontinuity.}

\item $\displaystyle f(x)=x^3+5x+1$,\ \ at\ \  $x=2$
\answer{Continuous}

\item $\displaystyle g(t)=\frac{t+1}{t^2+4}$,\ \ at\ \  $t=-1$
\answer{Continuous}

\item $\displaystyle h(y)=\frac{y^2+4y+4}{y+2}$,\ \ at\ \  $y=-2$
\answer{Removable Discontinuity}

\item $\displaystyle p(s)=\sqrt{s}-4$,\ \ at\ \  $s=2$
\answer{Continuous}

\item $\displaystyle f(x)=\left\{
\begin{array}{cl}
x^2+1,&\mbox{if}\ x\leq 1 \\ 
\frac{x+1}{x-1},& \mbox{if}\ x>1
\end{array}
\right.$,\ \ \ at \ $x=1$
\answer{Infinite Discontinuity}

\item $\displaystyle g(t)=\left\{ 
\begin{array}{cl}
2t+3,&\mbox{if} \ t\leq 2 \\ 
\frac{t^2-5t+6}{t-2},&\mbox{if} \ t>2
\end{array}
\right.$,\ \ \ at \ $t=2$
\answer{Jump Discontinuity}

\end{problemblock}

\item Let $c$ be a constant real number and $f$ be the function

	\[
		f(x)=
			\left\{
				\begin{array}{cc}
					\sqrt{-x}+1&\text{if }x<0\\
					x^2+c^2&\text{if }x\geq0\\
				\end{array}
			\right.
	\]
\begin{enumerate}
\item Explain why, for $c=-2$, the function $f$ is discontinuous at $x=0$.
\item Determine all real numbers $c$ for which $f$ is continuous at $x=0$.
\end{enumerate}
\answer{ 
\begin{enumerate}
\item $\displaystyle\lim_{x\to 0^-}f(x)=1$ and $\displaystyle\lim_{x\to 0^+}f(x)=4$.  Therefore $\displaystyle\lim_{x\to 0}f(x)$ does not exist.
\item $c=\pm 1$
\end{enumerate}
}

\item Where is the function $\displaystyle f(x)=\frac{x^2+3x+2}{x^2-1}$ continuous?
\answer{$\mathset{R}-\left\{-1,1\right\}$\,.  Note the limit actually exists at $x=-1$ but the function is not defined there.}

\item Using the Intermediate Value Theorem, show there is a real number $c$ strictly between $1$ and $3$ such that $c^3+2c^2=10$.  		
\answer{ Let $f(x)=x^3+2x^2$.  Then notice that $f(1)=3$ and $f(3)=33$.  Then since $f(1)<10<f(3)$ and $f$ is continuous, there exists a $c\in(1,3)$ such that $f(c)=10$ by the IVT.  Since $f(c)=c^3+2c^2$ the result follows.
}

\item Show that the equation $\displaystyle x^2+\cos x-2=0$ has a solution in the interval $(0,2)$.
\answer{$f(x)=x^2+\cos x-2$ is continuous on $[0,2]$, $f(0)=-1<0$, $f(2)\approx 1.58>0$, so by the Intermediate Value Theorem there is a $c$ in $(0,2)$ with $f(c)=0$.  i.e. $c^2+\cos c-2=0$ and $c$ is therefore a solution to the equation.}
