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%  Derivative Function                      %
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\problemgrouptitle{Derivative Function}

\item Which of the following graphs represent functions that are differentiable at $x=0$?  (Explain why or why not).
\begin{multicols}{4}
\begin{enumerate}
\item ~

\resizebox{0.87\columnwidth}{!}{
\asycode{%
import graph;
size(150,150,IgnoreAspect);
real f(real x) {return x^2-1;}
pair F(real x) {return (x,f(x));}
draw(graph(f,-1.5,1.5,operator ..),red+2);
xaxis("$x$",Arrow(HookHead,size=4),above=true);
yaxis("$y$",ymin=-1.2,Arrow(HookHead,size=4),above=true);
}
}

\item ~

\resizebox{0.87\columnwidth}{!}{
\asycode{%
import graph;
size(150,150,IgnoreAspect);
real f(real x) {return -x^3+.5;}
pair F(real x) {return (x,f(x));}
draw(graph(f,-1,1,operator ..),red+2);
xaxis("$x$",Arrow(HookHead,size=4),above=true);
yaxis("$y$",Arrow(HookHead,size=4),above=true);
}
}

\item ~

\resizebox{0.87\columnwidth}{!}{
\asycode{%
import graph;
size(150,150,IgnoreAspect);
real f(real x) {return -x+0.75;}
pair F(real x) {return (x,f(x));}
real g(real x) {return x^2+0.5;}
pair G(real x) {return (x,g(x));}
draw(graph(f,-.75,0,operator ..),red+2);
draw(graph(g,0,1,operator ..),red+2);
xaxis("$x$",Arrow(HookHead,size=4),above=true);
yaxis("$y$",Arrow(HookHead,size=4),above=true);
dot(F(0),4+black);
dot(F(0),3+white);
dot(G(0),4+black);
dot(G(0),3+red);
}
}

\item ~

\resizebox{0.87\columnwidth}{!}{
\asycode{%
import graph;
size(150,150,IgnoreAspect);
real f(real x) {return .5*x+.25;}
pair F(real x) {return (x,f(x));}
real g(real x) {return -x^2+0.25;}
pair G(real x) {return (x,g(x));}
draw(graph(f,-0.5,0,operator ..),red+2);
draw(graph(g,0,0.5,operator ..),red+2);
xaxis("$x$",xmax=0.7,Arrow(HookHead,size=4),above=true);
yaxis("$y$",ymax=0.4,Arrow(HookHead,size=4),above=true);
}
}
\end{enumerate}
\end{multicols}
\answer{ 
\begin{enumerate}
\item Differentiable
\item Differentiable
\item Not Differentiable.  Discontinuous at $x=0$.
\item Not differentiable.  At $x=0$ the left and right hand limits for the derivative are not equal:
%\begin{equation*}
\[\lim_{h\to0^-}\frac{f(x+h)-f(x)}{h}\neq \lim_{h\to0^+}\frac{f(x+h)-f(x)}{h},\]
%\end{equation*}
therefore the limit itself (the derivative) does not exist. Geometrically no tangent line may be drawn at the point so there can be no derivative as that is the tangent slope. 
\end{enumerate}
}
