CC Differentiability

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Section 2.3 Differentiability

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Motivating Questions

What does it mean graphically to say that a function $f$ is differentiable at $x = a ?$ 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?

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This section corresponds to 2.2 Definition of the Derivative in the workbook.

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In this section we aim to determine whether or not the function has a derivative

f^{'} (a)

x = a .

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Subsection 2.3.1 Being Differentiable at a Point

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We recall that a function

f

is said to be differentiable at

x = a

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)),

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)),

it is necessary for

f

to be continuous at

x = a :

f

fails to have a limit at

x = a,

f (a)

is not defined, or if

f (a)

does not equal the value of

lim_{x \to a} f (x),

then it doesn’t make sense to talk about a tangent line to the curve at this point.

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Indeed, it can be proved formally that if a function

f

is differentiable at

x = a,

then it must be continuous at

x = a .

Stated differently, if

f

is not continuous at

x = a,

then it is automatically the case that

f

is not differentiable there. For example, in Figure 2.3.1 below, both

f

and

g

fail to be differentiable at

x = 1

because neither function is continuous at

x = 1 .

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Figure 2.3.1. Functions $f$ and $g$ that are not continuous (and hence not differentiable) at $x = 1,$ and a function $h$ that is both continuous and differentiable at $x = 1 .$

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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,

where the graph of

y = f (x)

is displayed below in Figure 2.3.2. We notice that

f

has a sharp corner at the point

(1, 1),

and further observe that

f

is continuous at

x = 1

since

lim_{x \to 1} f (x) = 1 = f (1) .

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Figure 2.3.2. A function $f$ that is continuous at $x = 1$ but not differentiable at $x = 1;$ at right, we zoom in on the point $(1, 1)$ in a magnified version of the box shown in the left-hand plot.

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However, the function

f

in Figure 2.3.2 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

- 1

would be equally good candidates for the value of the derivative at

x = 1 .

Alternatively, we could use the limit definition of the derivative to attempt to compute

f^{'} (1),

and discover that the derivative does not exist. Finally, we can see visually in Figure 2.3.2 that this function does not have a tangent line at

x = 1 .

Regardless of how closely we examine the function by zooming in on

(1, 1)

on the graph of

y = f (x),

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.

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Subsection 2.3.2 Vertical Tangent Lines

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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 .

This means the tangent lines become very steep as we move closer to

x = a .

Example 2.3.3.

In this example, let

f (x) = \sqrt{x} .

In Figure 2.3.4 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,

we see that the tangent lines drawn become steeper and steeper, ultimately leading to a vertical tangent line at

x = 0 .

Figure 2.3.4. As we move closer to $x = 0,$ the tangent lines to the graph of $y = f (x)$ become steeper and steeper. Notice that the tangent line closest to $x = 0$ is nearly vertical.

We can also show this by calculating the limit of the derivative close to

x = 0

lim_{x \to 0} f^{'} (x) = lim_{x \to 0} \frac{1}{2 \sqrt{x}} = \infty .

Therefore,

f (x)

is not differentiable at

x = 0 .

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Subsection 2.3.3 Links Between Continuity, Differentiability, and Limits

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

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If $f$ is differentiable at $x = a,$ then $f$ is continuous at $x = a .$ Equivalently, if $f$ fails to be continuous at $x = a,$ then $f$ will not be differentiable at $x = a .$
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 $(a, f (a)) .$
If $f$ is differentiable at $x = a,$ then $f$ is locally linear at $x = a .$ In other words, a differentiable function looks linear when viewed up close because it resembles its tangent line at any given point of differentiability.

Example 2.3.5.

In this example, let

f

be the function whose graph is given below in Figure 2.3.6.

Figure 2.3.6. The graph of $y = f (x)$ for Example 2.3.5.

State all values of $a$ for which $f$ is not continuous at $x = a .$ For each, provide a reason for your conclusion.
State all values of $a$ for which $f$ is not differentiable at $x = a .$ For each, provide a reason for your conclusion.
State all values of $a$ for which $f$ is not differentiable, but is continuous at $x = a .$ Think about why this is the case.

Answer.

At $a = - 2,$ $lim_{x \to - 2} f (x)$ does not exist; at $a = - 1,$ $lim_{x \to - 1} f (x) \neq f (- 1);$ at $a = 2,$ $lim_{x \to 2} f (x)$ does not exist; at $a = 3,$ $f (3)$ is undefined.
$a = - 2, - 1, 2, 3,$ because $f$ is not continuous at these points; $a = - 3, 1,$ because $f$ does not have a tangent line at these points.
$a = - 3, 1 .$

Solution.

$f$ 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 $lim_{x \to - 1} f (x) = - 3.5,$ but $f (- 1) = 1 .$ $f$ is not continuous at $a = 3$ because $f (3)$ is not defined.
$f$ 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.
The only two points where $f$ is continuous but not differentiable are $a = - 3, 1 .$ This is because of the corner point (or cusp). These points fit the criteria for continuity, but there is no discernible tangent line.

Example 2.3.7.

True or false: if a function

p

is differentiable at

x = b,

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?

Answer.

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,

then

lim_{x \to b} p (x)

must exist” is true.

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Subsection 2.3.4 Summary

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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 $x = a .$ 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 $(a, f (a)) .$
Differentiability is a stronger condition than continuity, which is a stronger condition than having a limit. In particular, if $f$ is differentiable at $x = a,$ then $f$ is also continuous at $x = a,$ and if $f$ is continuous at $x = a,$ then $f$ has a limit at $x = a .$
A continuous function fails to be differentiable at any point where the graph has a corner point or cusp, or where the graph has a vertical tangent line.

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Exercises 2.3.5 Exercises

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1. Continuity and differentiability of a graph.

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Consider the function graphed below.

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At what

x

-values does the function appear to not be continuous?

x =

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At what

x

-values does the function appear to not be differentiable?

x =

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(Enter none if there are no $x$ -values that apply; enter $x$ -values as a comma-separated list, e.g., 1,3,5.)

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2. Differentiability of a graph.

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3. Differentiability of a graph.

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4. Continuity and differentiability of a graph.

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Consider the graph of the function

y = p (x)

that is provided in Figure 2.3.8. Assume that each portion of the graph of

p

is a straight line, as pictured.

Figure 2.3.8. At left, the piecewise linear function $y = p (x) .$ At right, axes for plotting $y = p^{'} (x) .$

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State all values of $a$ for which $lim_{x \to a} p (x)$ does not exist.
State all values of $a$ for which $p$ is not continuous at $a .$
State all values of $a$ for which $p$ is not differentiable at $x = a .$
On the axes provided in Figure 2.3.8, sketch an accurate graph of $y = p^{'} (x) .$

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5. Examples of functions.

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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.

A function $f$ that is continuous at $a = 2$ but not differentiable at $a = 2 .$
A function $g$ that is differentiable at $a = 3$ but does not have a limit at $a = 3 .$
A function $h$ that has a limit at $a = - 2,$ is defined at $a = - 2,$ but is not continuous at $a = - 2 .$
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

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6. Estimating the derivative at a point.

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Consider the function

g (x) = \sqrt{| x |} .

Use a graph to explain visually why $g$ is not differentiable at $x = 0 .$
Use the limit definition of the derivative to show that

$g^{'} (0) = lim_{h \to 0} \frac{\sqrt{| h |}}{h}$

.
Investigate the value of $g^{'} (0)$ by estimating the limit in (b) using small positive and negative values of $h .$ For instance, you might compute $\frac{\sqrt{| - 0.01 |}}{0.01} .$ 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 $g^{'} (0) ?$
Use your graph in (a) to sketch an approximate graph of $y = g^{'} (x) .$

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