
## Section1.8Differentiability

###### Motivating Questions
• What does it mean graphically to say that a function $f$ is differentiable at $x = a\text{?}$ How is this connected to the function being locally linear?

• How are the characteristics of a function having a limit, being continuous, and being differentiable at a given point related to one another?

In this section we aim to determine whether or not the function has a derivative $f'(a)$ at $x = a\text{.}$

### SubsectionBeing Differentiable at a Point

We recall that a function $f$ is said to be differentiable at $x = a$ if $f'(a)$ exists. Moreover, for $f'(a)$ to exist, we know that the graph of $y = f(x)$ must have a tangent line at the point $(a,f(a))\text{,}$ since the value of $f'(a)$ is precisely the slope of this line. Observe that in order to ask if $f$ has a tangent line at $(a,f(a))\text{,}$ it is necessary for $f$ to be continuous at $x = a\text{:}$ if $f$ fails to have a limit at $x = a\text{,}$ if $f(a)$ is not defined, or if $f(a)$ does not equal the value of $\lim_{x \to a} f(x)\text{,}$ then it doesn't make sense to talk about a tangent line to the curve at this point.

Indeed, it can be proved formally that if a function $f$ is differentiable at $x = a\text{,}$ then it must be continuous at $x = a\text{.}$ Stated differently, if $f$ is not continuous at $x = a\text{,}$ then it is automatically the case that $f$ is not differentiable there. For example, in Figure1.97 below, both $f$ and $g$ fail to be differentiable at $x = 1$ because neither function is continuous at $x = 1\text{.}$

A natural question to ask at this point is is there a difference between continuity and differentiability? In other words, can a function fail to be differentiable at a point where the function is continuous? To answer these questions, we consider a certain function $f\text{,}$ where the graph of $y=f(x)$ is displayed below in Figure1.98. We notice that $f$ has a sharp corner at the point $(1,1)\text{,}$ and further observe that $f$ is continuous at $x=1$ since $\lim_{x\to1} f(x)=1=f(1)\text{.}$

However, the function $f$ in Figure1.98 is not differentiable at $x = 1$ because $f'(1)$ fails to exist. One way to see this is to observe that $f'(x) = -1$ for every value of $x$ that is less than 1, while $f'(x) = +1$ for every value of $x$ that is greater than 1. That makes it seem that either $+1$ or $-1$ would be equally good candidates for the value of the derivative at $x = 1\text{.}$ Alternatively, we could use the limit definition of the derivative to attempt to compute $f'(1)\text{,}$ and discover that the derivative does not exist. Finally, we can see visually in Figure1.98 that this function does not have a tangent line at $x=1\text{.}$ Regardless of how closely we examine the function by zooming in on $(1,1)$ on the graph of $y=f(x)\text{,}$ it will always look like a V and never like a single line, which tells us there is no possibility for a tangent line there.

### SubsectionLocal Linearity

If a function does have a tangent line at a given point, then the function and the tangent line should appear essentially indistinguishable when we zoom in on the point of tangency.8For instance, see http://gvsu.edu/s/6J for an applet (due to David Austin, GVSU) where zooming in shows the increasing similarity between the tangent line and the curve. Conversely, if we zoom in on a point and the function looks like a single straight line, then the function should have a tangent line there and thus be differentiable at that point. Therefore, we say that a function that is differentiable at $x = a$ is locally linear. An example of this can be seen in Figure1.99 below, with the graph of a function $y=f(x)$ on the left and a magnified version with the tangent line to $f$ at $a$ on the right.

###### Example1.100

Let $g$ be the function given by the rule $g(x) = |x|\text{.}$

1. Visualize (or sketch) the graph of $y=g(x)\text{,}$ and use it to explain why $g$ is differentiable at every nonzero value of $x\text{.}$

2. Use the limit definition of the derivative to show that $g'(0) = \lim_{h \to 0} \frac{|h|}{h}\text{.}$ Then use left- and right- handed limits to explain why this limit does not exist. Conclude that $g$ is not differentiable at $x=0\text{.}$

Hint
1. What type of function is $g$ for all $x \lt 0\text{?}$ For all $x > 0\text{?}$

2. Recall that $g'(0) = \lim_{h \to 0} \frac{g(0 + h) - g(0)}{h}\text{.}$ What is the value of $|h|$ when $h \lt 0\text{?}$

1. $g$ is piecewise linear.

2. \begin{equation*} \lim_{h \to 0^-} \frac{|h|}{h}=-1, \ \text{and} \ \lim_{h\to0^+}\frac{|h|}h=1\text{.} \end{equation*}
Solution
1. We know that $g(x) = |x|$ is given by the formula $g(x) = -x$ when $x \lt 0$ and by $g(x) = x$ when $x \ge 0\text{.}$ Each of these pieces of $g$ is a straight line, so at every point other than the point where they meet, the function $g$ has a well-defined slope and thus is differentiable whenever $x\neq0\text{.}$ The graph of $y=g(x)$ is a translate of the earlier graph of $y=f(x)$ from Figure1.98.

2. Observe that

\begin{align*} g'(0) =\mathstrut \amp \lim_{h \to 0} \frac{g(0+h)-g(0)}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{|0+h|-|0|}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{|h|}{h} \end{align*}

To evaluate this limit, note that $|h| = h$ whenever $h > 0\text{,}$ and thus

\begin{equation*} \lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1\text{.} \end{equation*}

In contrast, we have $|h| = -h$ whenever $h \lt 0\text{,}$ and thus

\begin{equation*} \lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1\text{.} \end{equation*}

Since the right- and left-hand limits are not equal, it follows that

\begin{equation*} g'(0) = \lim_{h \to 0} \frac{|h|}{h} \end{equation*}

does not exist. Therefore, the function $g(x)=|x|$ is not differentiable at $x=0\text{.}$

### SubsectionVertical Tangent Lines

Another example of when a function can fail to be differentiable at a point $x=a$ is if the function has a vertical tangent at the point. In other words, when $f$ is continuous at $x=a$ and $\lim_{x\to a}|f'(x)|=\infty\text{.}$ This means the tangent lines become very steep as we move closer to $x=a\text{.}$

###### Example1.101

In this example, let $f(x)=\sqrt{x}\text{.}$

In Figure1.102 below, we have the graph of $y=f(x)$ along with a progression of tangent lines at points approaching $(0,0)$ on the graph. As we approach $x=0\text{,}$ we see that the tangent lines drawn become steeper and steeper, ultimately leading to a vertical tangent line at $x=0\text{.}$

We can also show this by calculating the limit of the derivative close to $x=0\text{:}$

\begin{equation*} \lim_{x\to 0}f'(x)=\lim_{x\to 0}\frac{1}{2\sqrt{x}}=\infty. \end{equation*}

Therefore, $f(x)$ is not differentiable at $x=0 \text{.}$

### SubsectionLinks Between Continuity, Differentiability, and Limits

To summarize the preceding discussion of differentiability, we make several important observations.

• If $f$ is differentiable at $x = a\text{,}$ then $f$ is continuous at $x = a\text{.}$ Equivalently, if$f$ fails to be continuous at $x = a\text{,}$ then $f$ will not be differentiable at $x = a\text{.}$

• A function can be continuous at a point without being differentiable there. In particular, a function $f$ is not differentiable at $x = a$ if the graph has a sharp corner (or cusp) at the point $(a,f(a))\text{.}$

• If $f$ is differentiable at $x = a\text{,}$ then $f$ is locally linear at $x = a\text{.}$ In other words, a differentiable function looks linear when viewed up close because it resembles its tangent line at any given point of differentiability.

###### Example1.103

In this example, let $f$ be the function whose graph is given below in Figure1.104.

1. State all values of $a$ for which $f$ is not continuous at $x = a\text{.}$ For each, provide a reason for your conclusion.

2. State all values of $a$ for which $f$ is not differentiable at $x = a\text{.}$ For each, provide a reason for your conclusion.

3. State all values of $a$ for which $f$ is not differentiable, but is continuous at $x = a\text{.}$ Think about why this is the case.

Hint
1. You might start by looking for places you would lift your pencil when drawing the graph, identifying points where $f$ is not continuous. Calculating limits may also be useful.

2. You might start by identifying points where $f$ is not continuous.

3. Thinking back to Example1.100 may be useful.

1. At $a = -2 \text{,}$ $\lim_{x\to-2}f(x)$ does not exist; at $a=-1\text{,}$ $\lim_{x\to-1}f(x)\neq f(-1)\text{;}$ at $a=2\text{,}$ $\lim_{x\to2}f(x)$ does not exist; at $a=3\text{,}$ $f(3)$ is undefined.

2. $a = -2, -1, 2, 3\text{,}$ because $f$ is not continuous at these points; $a=-3,1\text{,}$ because $f$ does not have a tangent line at these points.

3. $a = -3,1\text{.}$

Solution
1. $f$ is not continuous at $a = -2, 2$ because at each of these points $\lim_{x\to a}f(x)$ does not exist. $f$ is not continuous at $a = -3$ because $\lim_{x\to -1}f(x)=-3.5\text{,}$ but $f(-1)=1\text{.}$ $f$ is not continuous at $a=3$ because $f(3)$ is not defined.

2. $f$ is not differentiable at $a = -2, -1, 2, 3$ because at each of these points $f$ is not continuous. In addition, $f$ is not differentiable at $a = -3$ and $a = 1$ because the graph of $f$ has a corner point (or cusp) at each of these values.

3. The only two points where $f$ is continuous but not differentiable are $a=-3,1\text{.}$ This is because of the corner point (or cusp). These points fit the criteria for continuity, but there is no discernible tangent line.

###### Example1.105

True or false: if a function $p$ is differentiable at $x = b\text{,}$ then $\lim_{x \to b} p(x)$ must exist. Write at least one sentence to justify your choice.

Hint

What does being differentiable at a point tell you about continuity there?

True.

Solution

We know that a function $f$ is continuous whenever it is differentiable, and that one characteristic of $f$ being continuous at $x=a$ is that $\lim_{x\to a}f(x)$ exists. Therefore the statement if a function $p$ is differentiable at $x = b\text{,}$ then $\lim_{x \to b} p(x)$ must exist is true.

### SubsectionSummary

• A function $f$ is differentiable at $x = a$ whenever $f'(a)$ exists, which means that $f$ has a tangent line at $(a,f(a))$ and thus $f$ is locally linear at $x = a\text{.}$ Informally, this means that the function looks like a line when viewed up close at $(a,f(a))$ and that there is not a corner point or cusp at $(a,f(a))\text{.}$

• Differentiability is a stronger condition than continuity, which is a stronger condition than having a limit. In particular, if $f$ is differentiable at $x = a\text{,}$ then $f$ is also continuous at $x = a\text{,}$ and if $f$ is continuous at $x = a\text{,}$ then $f$ has a limit at $x = a\text{.}$

### SubsectionExercises

Consider the graph of the function $y = p(x)$ that is provided in Figure1.106. Assume that each portion of the graph of $p$ is a straight line, as pictured.

1. State all values of $a$ for which $\lim_{x \to a} p(x)$ does not exist.

2. State all values of $a$ for which $p$ is not continuous at $a\text{.}$

3. State all values of $a$ for which $p$ is not differentiable at $x = a\text{.}$

4. On the axes provided in Figure1.106, sketch an accurate graph of $y = p'(x)\text{.}$

For each of the following prompts, give an example of a function that satisfies the stated criteria; a formula or a graph, with reasoning, is sufficient for each. If no such example is possible, explain why.

1. A function $f$ that is continuous at $a = 2$ but not differentiable at $a = 2\text{.}$

2. A function $g$ that is differentiable at $a = 3$ but does not have a limit at $a=3\text{.}$

3. A function $h$ that has a limit at $a = -2\text{,}$ is defined at $a = -2\text{,}$ but is not continuous at $a = -2\text{.}$

4. A function $p$ that satisfies all of the following:

• $p(-1) = 3$ and $\lim_{x \to -1} p(x) = 2$

• $p(0) = 1$ and $p'(0) = 0$

• $\lim_{x \to 1} p(x) = p(1)$ and $p'(1)$ does not exist

Consider the function $g(x) = \sqrt{|x|}\text{.}$

1. Use a graph to explain visually why $g$ is not differentiable at $x = 0\text{.}$

2. Use the limit definition of the derivative to show that

\begin{equation*} g'(0) = \lim_{h \to 0} \frac{\sqrt{|h|}}{h}\text{.} \end{equation*}
3. Investigate the value of $g'(0)$ by estimating the limit in (b) using small positive and negative values of $h\text{.}$ For instance, you might compute $\frac{\sqrt{|-0.01|}}{0.01}\text{.}$ Be sure to use several different values of $h$ (both positive and negative), including ones closer to 0 than 0.01. What do your results tell you about $g'(0)\text{?}$

4. Use your graph in (a) to sketch an approximate graph of $y = g'(x)\text{.}$