
## Section10.1Differentiable Functions

### Subsection10.1.1Differentiability

A function is differentiable at points where the derivative is defined. Alternatively, because the derivative at a point represents the slope of the tangent line, we say the function is differentiable at a point wherever the function has a well-defined tangent line.

###### Definition10.1.1Differentiability

A function $f$ is differentiable at $a$ if $f'(a)$ exists, or more precisely the limit

\begin{equation*} \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \end{equation*}

exists.

A function is not differentiable if the limit does not exist. There are several reasons this might occur. The first reason is if the function is not continuous.

Suppose that $f$ is differentiable at $a\text{.}$ This means that $f'(a)$ is a value defined by

\begin{equation*} \lim_{x \to a}\frac{f(x)-f(a)}{x-a} = f'(a). \end{equation*}

We also know that

\begin{equation*} \lim_{x \to a}[x-a] = a-a = 0. \end{equation*}

Using the product rule (LC:Product), this implies

\begin{equation*} \lim_{x \to a} [f(x)-f(a)] = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \cdot (x-a) = f'(a) \cdot 0 = 0. \end{equation*}

Because $\displaystyle \lim_{x \to a} f(a) = f(a)$ (LE:Constant), we know that

\begin{equation*} \lim_{x \to a} f(x) = \lim_{x \to a} [f(x)-f(a) + f(a)] = 0 + f(a) = f(a) \end{equation*}

using the sum rule (LC:Sum). Therefore, $f$ is continuous at $a\text{.}$

Another way that a function might not have be differentiable is where it is continuous but has a corner. This means that the slope at the point looks different from either of the two sides. Mathematically, if we computed the one-sided limits for the formula of the derivative, we would get two different values.

###### Example10.1.3

Consider the piecewise function defined by

\begin{equation*} f(x) = \begin{cases} x^2, & x \le 1, \\ x, & x \gt 1.\end{cases} \end{equation*}

Determine if $f$ is differentiable at $x=1\text{.}$

Solution

This function is continuous because the limit on the left and the limit on the right are equal to the value of the function at $x=1\text{,}$ as follows:

\begin{align*} \lim_{x \to 1^-}f(x) &= \lim_{x \to 1}x^2 = 1^2 = 1,\\ \lim_{x \to 1^+}f(x) &= \lim_{x \to 1}x = 1 = 1,\\ f(1) &= 1^2 = 1. \end{align*}

Now that we know the function is continuous, we can think about the functions to the left and right, $f_\ell(x)=x^2$ and $f_r(x)=x\text{,}$ respectively. Because $f_\ell'(x)=2x\text{,}$ we know that

\begin{equation*} \lim_{h \to 0^-} \frac{f(1+h)-f(1)}{h} = f_\ell'(1) = 2. \end{equation*}

Similarly, because $f_r'(x)=1\text{,}$ we know that

\begin{equation*} \lim_{h \to 0^+} \frac{f(1+h)-f(1)}{h} = f_r'(1)=1. \end{equation*}

Since the left and right limits are different, we conclude that $\displaystyle \lim_{h \to 0}\frac{f(1+h)-f(1)}{h}$ does not exist and $f$ is not differentiable at $x=1\text{.}$ The figure below illustrates this graph, showing that there is a corner at $x=1\text{.}$

###### Example10.1.4

Consider the piecewise function defined by

\begin{equation*} f(x) = \begin{cases} x^2-3x+8 & x \lt 2, \\ 5x-x^2, & x \ge 2.\end{cases} \end{equation*}

Determine if $f$ is differentiable at $x=2\text{.}$

Solution

This function is continuous because the limit on the left and the limit on the right are equal to the value of the function at $x=2\text{,}$ as follows:

\begin{align*} \lim_{x \to 2^-}f(x) &= \lim_{x \to 2}(x^2-3x+8) = 2^2-3(2)+8 = 6,\\ \lim_{x \to 2^+}f(x) &= \lim_{x \to 2}(5x-x^2) = 5(2)-2^2 = 6,\\ f(2) &= 5(2)-2^2 = 6. \end{align*}

Now that we know the function is continuous, we can think about the functions to the left and right, $f_\ell(x)=x^2-3x+8$ and $f_r(x)=5x-x^2\text{,}$ respectively. Because $f_\ell'(x)=2x-3\text{,}$ we know that

\begin{equation*} \lim_{h \to 0^-} \frac{f(2+h)-f(2)}{h} = f_\ell'(2) = 2(2)-3=1. \end{equation*}

Similarly, because $f_r'(x)=5-2x\text{,}$ we know that

\begin{equation*} \lim_{h \to 0^+} \frac{f(2+h)-f(2)}{h} = f_r'(2)=5-2(2)=1. \end{equation*}

Since the left and right limits are the same, we conclude that $\displaystyle \lim_{h \to 0}\frac{f(2+h)-f(2)}{h} = 1$ and $f'(2)=1\text{.}$ So $f$ is differentiable at $x=2\text{.}$ The graph of this function has the two parabolas join smoothly with no corner at $x=2\text{.}$

### Subsection10.1.2Consequences of Differentiability

There are a number of important consequences of a function being differentiable. These consequences are stated as mathematical theorems that you will need to know by name. We begin by introducing terminology about local extreme values.

###### Definition10.1.5Local Maximum and Minimum

A function $f$ has a local maximum at a point $x=a$ if $f(a) \ge f(x)$ for all $x$ in a neighborhood of $a\text{.}$ It has a local minimum at $x=a$ if $f(a) \le f(x)$ for all $x$ in a neighborhood of $a\text{.}$

The first theorem is about the slope at a local extreme. It guarantees that a local extreme can only occur where the function either is not differentiable or has a horizontal tangent line.

The second theorem combines the Extreme Value Theorem with Fermat's Theorem. If a function is continuous on a closed interval $[a,b]\text{,}$ then it must achieve both a maximum and a minimum value. If that function has $f(a)=f(b)\text{,}$ then one of the extreme values must occur inside the interval at some point $c \in (a,b)\text{.}$ If the function is also differentiable, then we must have $f'(c)=0\text{.}$ This result is named Rolle's theorem.

The consequence of Rolle's theorem is that if a function starts and ends at the same value over an interval, it must turn around somewhere. For a differentiable function, the slope at that point must be $f'(c)=0\text{.}$

The third theorem about differentiability applies Rolle's theorem to say something about the average rate of change. Recall that the average rate of change,

\begin{equation*} \left.\frac{\Delta f}{\Delta x}\right|_{[a,b]} = \frac{f(b)-f(a)}{b-a}, \end{equation*}

is the slope of the line, called a secant line, that joins the points $(a,f(a))$ and $(b,f(b))\text{.}$ The Mean Value Theorem guarantees that a continuous and differentiable function will have some point at which the tangent line has the same slope as the secant line over the interval.

Let $s(x)$ be the linear function corresponding to this secant line and then define $g(x)=f(x)-s(x)\text{.}$ Since $s(a)=f(a)$ and $s(b)=f(b)\text{,}$ we have $g(a)=g(b)=0\text{.}$ If $f$ is continuous and differentiable, then so is $g\text{.}$ Rolle's theorem guarantees that $g'(c)=f'(c)-s'(c) = 0$ for some value $c \in (a,b)\text{.}$ Thus, $f'(c)=\left.\frac{\Delta f}{\Delta x}\right|_{[a,b]}\text{.}$