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\chead{\Large Differentiable Functions}
\rfoot{\footnotesize Last revised: October 3, 2017}
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Recall the definition of continuous functions.\vspace{0.07in}

\textbf{Definition:} (Continuity at a point) \ A function $y=f\left(
x\right) $ is \textbf{continuous at a number }$\boldsymbol{c}$ of its domain
if the two-sided limit exists and $\lim\limits_{x\rightarrow c}f\left(
x\right) =f\left( c\right) $.\vspace{0.07in}

\textbf{Definition:} \ (Continuity on an interval)

(\textit{Open Interval}) \ A function $y=f\left( x\right) $ is continuous on
an interval $\left( a,b\right) $ if it is continuous at every $c$ in $\left(
a,b\right) $.\newline
(\textit{Closed Interval}) \ A function $y=f\left( x\right) $ is continuous
on an interval $\left[ a,b\right] $ if it is continuous at every $c$ in $%
\left( a,b\right) $ and 
\begin{equation*}
\lim\limits_{x\rightarrow a^{+}}f\left( x\right) =f\left( a\right) \text{ \
\ or \ \ }\lim\limits_{x\rightarrow b^{-}}f\left( x\right) =f\left( b\right) 
\text{, \ respectively.}
\end{equation*}

End-points of the interval require only one-sided limits.\vspace{0.07in}

Another way to express continuity is to say that $\lim\limits_{h\rightarrow
0}f\left( x+h\right) =f\left( x\right) $. \ Another alternative statement of
continuity is $\lim\limits_{x\rightarrow c}f\left( x\right) =f\left(
\lim\limits_{x\rightarrow c}x\right) $, so that there is a commutativity
between taking the limit and taking the function values.\vspace{0.07in}

\textbf{Definition:} \ Suppose that $f$ is a function and $c$ is an interior
point of its domain. \ If the (two-sided) limit%
\begin{equation*}
\lim\limits_{h\rightarrow 0}\dfrac{f\left( c+h\right) -f\left( c\right) }{h}
\end{equation*}%
exists (and is finite) we say that $f$ is \textbf{differentiable at }$%
\boldsymbol{c}$ and denote this limit as $f^{\prime }\left( c\right) $.%
\vspace{0.07in}

\textbf{Theorem:} \ If $f$ is differentiable at $a$, then it is continuous
there.\vspace{0.07in}

Proof: Suppose that $f$ is differentiable at a number $a$. \ Then $f^{\prime
}\left( a\right) $ exists which means that $f\left( a\right) $ exists and
the limit $\lim\limits_{h\rightarrow 0}\dfrac{f\left( a+h\right) -f\left(
a\right) }{h}$ also exists and is finite. \ Let us start with the true
statement that $0=0\cdot f^{\prime }\left( a\right) $.%
\begin{eqnarray*}
0 &=&0\cdot f^{\prime }\left( a\right) \\
0 &=&\lim\limits_{h\rightarrow 0}h\cdot \lim\limits_{h\rightarrow 0}\dfrac{%
f\left( a+h\right) -f\left( a\right) }{h}\text{ \ \ \ \ \ \ \ \ \ \ \ by the
product rule of limits} \\
0 &=&\lim\limits_{h\rightarrow 0}\left( h\cdot \dfrac{f\left( a+h\right)
-f\left( a\right) }{h}\right) \text{ \ \ \ \ \ \ \ \ \ \ \ cancel out }h \\
0 &=&\lim\limits_{h\rightarrow 0}\left( f\left( a+h\right) -f\left( a\right)
\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by the difference rule
of limits} \\
0 &=&\lim\limits_{h\rightarrow 0}f\left( a+h\right)
-\lim\limits_{h\rightarrow 0}f\left( a\right) \\
\lim\limits_{h\rightarrow 0}f\left( a\right) &=&\lim\limits_{h\rightarrow
0}f\left( a+h\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ by the constant rule of limits} \\
f\left( a\right) &=&\lim\limits_{h\rightarrow 0}f\left( a+h\right)
\end{eqnarray*}%
and $f\left( a\right) =\lim\limits_{h\rightarrow 0}f\left( a+h\right) $
means that $f$ is continuous at $a$.\vspace{0.09in}

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So, differentiability implies continuity. \ What about backwards? \ Are
continuous functions necessarily differentiable? \ The answer is no. \
Consider the function\vspace{0.09in}\vspace{0.09in} \newline
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\ \ \ \ \ \ \ \ \ $f\left( x\right) =\left\vert x\right\vert =\left\{ 
\begin{array}{ccc}
x & \text{if} & x\geq 0 \\ 
-x & \text{if} & x<0%
\end{array}%
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\ Although this function is continuous at zero, it is not differentiable
there. \ Recall that the derivative is a two-sided limit. \ As $h$
approaches zero, it is negative when we compute the left-limit and positive
when we compute the right limit.%
\begin{eqnarray*}
\lim\limits_{h\rightarrow 0^{-}}\dfrac{f\left( 0+h\right) -f\left( 0\right) 
}{h} &=&\lim\limits_{h\rightarrow 0^{-}}\dfrac{f\left( h\right) -f\left(
0\right) }{h}=\lim\limits_{h\rightarrow 0^{-}}\dfrac{-h-0}{h}%
=\lim\limits_{h\rightarrow 0^{-}}-1=-1\text{ \ \ \ and} \\
\lim\limits_{h\rightarrow 0^{+}}\dfrac{f\left( 0+h\right) -f\left( 0\right) 
}{h} &=&\lim\limits_{h\rightarrow 0^{+}}\dfrac{f\left( h\right) -f\left(
0\right) }{h}=\lim\limits_{h\rightarrow 0^{+}}\dfrac{h-0}{h}%
=\lim\limits_{h\rightarrow 0^{+}}1=1
\end{eqnarray*}%
So the derivative, a two-sided limit, is not defined at $x=0$.\vspace{0.07in}%
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\vspace{0.09in}\vspace{0.09in} \newline
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\ \ \ \ \ \ $f^{\prime }\left( x\right) =\left\{ 
\begin{array}{ccc}
1 & \text{if} & x>0 \\ 
\text{undefined} & \text{if} & x=0 \\ 
-1 & \text{if} & x<0%
\end{array}%
\right. $

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When the function $f$ is continuous but not differentiable because the
left-hand side derivative and the right-hand side derivative exist and are
finite but not equal, the graph has a spike there. \ For example, the graph
of $f\left( x\right) =\left\vert x\right\vert $ has a spike at $x=0$. \
Differentiable functions have graphs that are continuous and smooth. \ 

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