MA005: Calculus I

The contrapositive form of this theorem tells about some functions which do not have derivatives:

Proof of the Theorem: We assume that the hypothesis, $f$ is differentiable at the point $c$, is true so 

$\lim\limits_{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$ exists and equals $\mathrm{f}^{\prime}(\mathrm{c})$. We want to show that $\mathrm{f}$ must necessarily be continuous at $\mathrm{c}$ $: \lim\limits_{h \rightarrow 0} f(c+h)=f(c)$

Since $\mathrm{f}(\mathrm{c}+\mathrm{h})$ can be written as $f(c+h)=f(c)+\left\{\frac{f(c+h)-f(c)}{h}\right\} \cdot h$, we have

$\begin{aligned} &\lim\limits_{h \rightarrow 0} f(c+h)=\lim\limits_{h \rightarrow 0}\left(f(c)+\left\{\frac{f(c+h)-f(c)}{h}\right\} \cdot h\right) \\ &=\lim\limits_{h \rightarrow 0} f(c)+\lim\limits_{h \rightarrow 0}\left\{\frac{f(c+h)-f(c)}{h}\right\} \cdot \lim _{h \rightarrow 0}(h)=\mathrm{f}(\mathrm{c})+\mathrm{f}^{\prime}(\mathrm{c}) \cdot 0=\mathrm{f}(\mathrm{c}) \end{aligned}$

Therefore $f$ is continuous at $c$. 

It is important to clearly understand what is meant by this theorem and what is not meant: If the function is differentiable at a point, then the function is automatically continuous at that point. If the function is continuous at a point, then the function may or may not have a derivative at that point.  

If the function is not continuous at a point, then the function is not differentiable at that point. 

Example 1: Show that $\mathrm{f}(\mathrm{x})=[\mathrm{x}]=\mathrm{INT}(\mathrm{x})$ is not continuous and not differentiable at 2 (Fig. 1).

Solution: The one-sided limits, $\lim\limits_{x \rightarrow 2^{+}} \operatorname{INT}(\mathrm{x})=2$ and $\lim\limits_{x \rightarrow 2^{-}} \operatorname{INT}(x)=1$, have different values so $\lim\limits_{x \rightarrow 2} \operatorname{INT}(x)$ does not exist, and $\operatorname{INT}(\mathrm{x})$ is not continuous at 2. Since $\mathrm{f}(\mathrm{x})=\mathrm{INT}(\mathrm{x})$ is not continuous at 2, it is not differentiable there.

Lack of continuity is enough to imply lack of differentiability, but the next two examples show that continuity is not enough to guarantee differentiability. 

Example 2: Show that $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$ is continuous but not differentiable at $\mathrm{x}=0$ (Fig. 2)

Solution: $\lim\limits_{x \rightarrow 0}|\mathrm{x}|=0=|0|$ so $\mathrm{f}$ is continuous at $0$, but we showed in Section 2.1 that the absolute value function was not differentiable at $\mathrm{x}=0$.

A function is not differentiable at a cusp or a "corner".

Example 3: Show that $\mathrm{f}(\mathrm{x})=\sqrt[3]{\mathrm{x}}=\mathrm{x}^{1 / 3}$ is continuous but not differentiable at $\mathrm{x}=0$ (Fig. 3)

Solution: $\lim\limits_{x \rightarrow 0^{-}} \sqrt[3]{x}=\lim\limits_{x \rightarrow 0^{+}} \sqrt[3]{x}=0$ so $\lim\limits_{x \rightarrow 0} \sqrt[3]{x}=0=\sqrt{0}$, and $\mathrm{f}$ is continuous at $0$. $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{3} \mathrm{x}^{-(2 / 3)}=\frac{1}{3 \mathrm{x}^{2 / 3}}$ which is undefined at $\mathrm{x}=0$ so $\mathrm{f}$ is not differentiable at $0$.

A function is not differentiable where its tangent line is vertical.

Practice 1: At which integer values of $x$ is the graph of f in Fig. 4 continuous?  differentiable? 

Graphically, a function is continuous if and only if its graph is connected and does not have any holes or breaks. 

Graphically, a function is differentiable if and only if it is continuous and its graph is smooth with no  corners or vertical tangent lines.