2.6 Differentiability & continuity

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2. Derivative basics

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Differentiability & continuity

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What you’ll learn

How to find where a function is not differentiable
The relationship between continuity and differentiability

What is differentiability?

A function $f (x)$ is differentiable at $x = a$ if its derivative exists at that point.

There are four common cases where a function is not differentiable at a point $x = a$ :

1. Discontinuities

A function must be continuous to be differentiable. If there is a removable, jump, or infinite discontinuity at $x = a$ , then the derivative does not exist at the point.

Often seen in rational, piecewise, or logarithmic functions, which may have discontinuities and/or asymptotes
e.g. $f (x) = \frac{1}{x}$

2. Sharp corners

The function changes direction abruptly. The one-sided derivatives exist, but they are not equal.

Often seen in absolute value or piecewise functions
e.g. $f (x) = ∣ x ∣$

3. Cusps

Cusps are like corners, but the slope becomes unbounded: one or both one-sided derivatives approach infinity, and they approach opposite infinities.

Often seen in power functions with fractional exponents
e.g. $f (x) = x^{2/3}$

4. Vertical tangents

The tangent line becomes vertical. The one-sided derivatives both approach infinity with the same sign.

Often seen in radical functions
e.g. $f (x) = x^{1/3}$

In the last three cases, the function is continuous, but the graph is not smooth at the non-differentiable point (for a visual, graph each in Desmos).

Remember this phrase:

$"If differentiable, then continuous." (D \Rightarrow C)$

Consequently, the contrapositive is always true:

If not continuous at $x = a$ , then not differentiable there $(not C \Rightarrow not D)$ .

However, the converse and inverse are not always true. A continuous function can still fail to be differentiable at certain points, e.g. $∣ x ∣$ at $x = 0$ .

Checking for differentiability

To check if $f (x)$ is differentiable at $x = a$ ,

Continuity: First check whether $f (x)$ is continuous by verifying the conditions stated in Section 1.6: Continuity. If it’s not continuous at $x = a$ , then it’s not differentiable.
Differentiate: Find $f^{'} (x)$ and look for any $x$ -values where $f^{'} (x)$ is undefined.
If necessary (mostly for piecewise functions): Check that the one-sided derivatives at $x = a$ both exist, are finite, and match. In other words, there must be a finite value $L$ such that

$x \to a^{-} lim \frac{f ( x ) - f ( a )}{x - a} = L = x \to a^{+} lim \frac{f ( x ) - f ( a )}{x - a}$

Example 1: From functions

For each of the following functions, find the $x$ -value(s) at which it is non-differentiable.

a) $a (x) = \frac{1}{x + 1}$

b) $b (x) = ∣ x - 1∣$

c) $c (x) = x^{2/3} + 1$

d) $d (x) = x^{1/3}$

Solutions

a) $a (x) = \frac{1}{x + 1}$

(spoiler)

The function has a vertical asymptote at $x = - 1$ and is discontinuous. A function that is not continuous is not differentiable at the point, and the derivative does not exist at $x = - 1$ .

b) $b (x) = ∣ x - 1∣$

(spoiler)

$b (x)$ is continuous for all values of $x$ . To analyze the derivative, first rewrite as a piecewise function:

$b (x) = ⎩ ⎨ ⎧ x - 1 - (x - 1) if x \geq 1 if x < 1$

Then differentiate each piece:

$b (x) = ⎩ ⎨ ⎧ 1 - 1 if x > 1 if x < 1$

Note that the equality is dropped because the derivative is not defined at $x = 1$ .

As $x \to 1^{+}$ , the derivative approaches $1$ , and as $x \to 1^{-}$ , it approaches $- 1$ . Because these one-sided derivatives don’t match, $b^{'} (1)$ does not exist. On the graph, this shows up as a sharp corner at $x = 1$ .

c) $c (x) = x^{2/3} + 1$

(spoiler)

$c (x)$ is continuous with derivative

$c^{'} (x) = \frac{2}{3} x^{- 1/3}$

$= \frac{2}{3 x ^{1/3}}$

This derivative is undefined when the denominator is $0$ , which happens at $x = 0$ .

Also, notice the one-sided behavior:

As $x \to 0^{-}$ , $c^{'} (x) \to - \infty$
As $x \to 0^{+}$ , $c^{'} (x) \to \infty$

Since the one-sided derivatives approach opposite infinities, the graph has a cusp at $x = 0$ , and the function is not differentiable at that point.

d) $d (x) = x^{1/3}$

(spoiler)

Differentiating

$d^{'} (x) = \frac{1}{3} x^{- 2/3} = \frac{1}{3 3 x ^{2}}$

which is undefined at $x = 0$ .

Example 2: From a graph

The graph of $f (x)$ shown below has a horizontal tangent at $x = b$ and a vertical tangent at $x = d$ . At which of the values $a, b, c,$ and $d$ is $f (x)$ not differentiable?

Solution

(spoiler)

At $x = a$ , the function is continuous but the one-sided derivatives are not equal, so $f$ is not differentiable at the point.

At $x = b$ , the tangent line is horizontal so $f^{'} (b) = 0$ , and $f$ is differentiable.

At $x = c$ , the function has a jump discontinuity, so it is not differentiable at the point.

At $x = d$ , the vertical tangent line has a slope that is undefined, so $f^{'} (d)$ does not exist and is not differentiable at that point.

Example 3: Finding unknown variables

Consider the function $f (x)$ defined as:

$f (x) = ⎩ ⎨ ⎧ 3 x + 2 a x^{2} - b x if x < 1 if x \geq 1$

Find the values of $a$ and $b$ such that $f (x)$ is differentiable for all values of $x$ .

(spoiler)

Answer: $a = - 2, b = - 7$

If a piecewise function is differentiable, then it must be continuous.

For $f (x)$ to be continuous at $x = 1$ , both $f (1)$ and $x \to 1 lim f (x)$ must exist and be equal.

First,

$f (1) = a (1)^{2} - b (1) = a - b$

Next, compute the one-sided limits:

Left:

$x \to 1^{-} lim (3 x + 2) = 5$

Right:

$x \to 1^{+} lim (a x^{2} - b x) = a - b$

So continuity requires $a - b = 5$ .

Since there are two unknowns, we need a second equation. Differentiate each piece (treating $a$ and $b$ as constants):

$f^{'} (x) = ⎩ ⎨ ⎧ 3 2 a x - b if x < 1 if x > 1$

For differentiability at $x = 1$ , the one-sided derivatives must match:

$3 = 2 a (1) - b$

Now solve the system:

$a - b 2 a - b = 5 = 3$

which gives

$a = - 2 b = - 7$

Graph

$f (x) = ⎩ ⎨ ⎧ 3 x + 2 2 x^{2} + 7 x if x < 1 if x \geq 1$

in Desmos to confirm that the curve is unbroken and also smooth.

Differentiability & continuity

What you’ll learn

How to find where a function is not differentiable
The relationship between continuity and differentiability

What is differentiability?

A function $f (x)$ is differentiable at $x = a$ if its derivative exists at that point.

There are four common cases where a function is not differentiable at a point $x = a$ :

1. Discontinuities

A function must be continuous to be differentiable. If there is a removable, jump, or infinite discontinuity at $x = a$ , then the derivative does not exist at the point.

Often seen in rational, piecewise, or logarithmic functions, which may have discontinuities and/or asymptotes
e.g. $f (x) = \frac{1}{x}$

2. Sharp corners

The function changes direction abruptly. The one-sided derivatives exist, but they are not equal.

Often seen in absolute value or piecewise functions
e.g. $f (x) = ∣ x ∣$

3. Cusps

Cusps are like corners, but the slope becomes unbounded: one or both one-sided derivatives approach infinity, and they approach opposite infinities.

Often seen in power functions with fractional exponents
e.g. $f (x) = x^{2/3}$

4. Vertical tangents

The tangent line becomes vertical. The one-sided derivatives both approach infinity with the same sign.

Often seen in radical functions
e.g. $f (x) = x^{1/3}$

In the last three cases, the function is continuous, but the graph is not smooth at the non-differentiable point (for a visual, graph each in Desmos).

Remember this phrase:

$"If differentiable, then continuous." (D \Rightarrow C)$

Consequently, the contrapositive is always true:

If not continuous at $x = a$ , then not differentiable there $(not C \Rightarrow not D)$ .

However, the converse and inverse are not always true. A continuous function can still fail to be differentiable at certain points, e.g. $∣ x ∣$ at $x = 0$ .

Checking for differentiability

To check if $f (x)$ is differentiable at $x = a$ ,

Continuity: First check whether $f (x)$ is continuous by verifying the conditions stated in Section 1.6: Continuity. If it’s not continuous at $x = a$ , then it’s not differentiable.
Differentiate: Find $f^{'} (x)$ and look for any $x$ -values where $f^{'} (x)$ is undefined.
If necessary (mostly for piecewise functions): Check that the one-sided derivatives at $x = a$ both exist, are finite, and match. In other words, there must be a finite value $L$ such that

$x \to a^{-} lim \frac{f ( x ) - f ( a )}{x - a} = L = x \to a^{+} lim \frac{f ( x ) - f ( a )}{x - a}$

Example 1: From functions

For each of the following functions, find the $x$ -value(s) at which it is non-differentiable.

a) $a (x) = \frac{1}{x + 1}$

b) $b (x) = ∣ x - 1∣$

c) $c (x) = x^{2/3} + 1$

d) $d (x) = x^{1/3}$

Solutions

a) $a (x) = \frac{1}{x + 1}$

(spoiler)

The function has a vertical asymptote at $x = - 1$ and is discontinuous. A function that is not continuous is not differentiable at the point, and the derivative does not exist at $x = - 1$ .

b) $b (x) = ∣ x - 1∣$

(spoiler)

$b (x)$ is continuous for all values of $x$ . To analyze the derivative, first rewrite as a piecewise function:

$b (x) = ⎩ ⎨ ⎧ x - 1 - (x - 1) if x \geq 1 if x < 1$

Then differentiate each piece:

$b (x) = ⎩ ⎨ ⎧ 1 - 1 if x > 1 if x < 1$

Note that the equality is dropped because the derivative is not defined at $x = 1$ .

c) $c (x) = x^{2/3} + 1$

(spoiler)

$c (x)$ is continuous with derivative

$c^{'} (x) = \frac{2}{3} x^{- 1/3}$

$= \frac{2}{3 x ^{1/3}}$

This derivative is undefined when the denominator is $0$ , which happens at $x = 0$ .

Also, notice the one-sided behavior:

As $x \to 0^{-}$ , $c^{'} (x) \to - \infty$
As $x \to 0^{+}$ , $c^{'} (x) \to \infty$

Since the one-sided derivatives approach opposite infinities, the graph has a cusp at $x = 0$ , and the function is not differentiable at that point.

d) $d (x) = x^{1/3}$

(spoiler)

Differentiating

$d^{'} (x) = \frac{1}{3} x^{- 2/3} = \frac{1}{3 3 x ^{2}}$

which is undefined at $x = 0$ .

Example 2: From a graph

The graph of $f (x)$ shown below has a horizontal tangent at $x = b$ and a vertical tangent at $x = d$ . At which of the values $a, b, c,$ and $d$ is $f (x)$ not differentiable?

Solution

(spoiler)

At $x = a$ , the function is continuous but the one-sided derivatives are not equal, so $f$ is not differentiable at the point.

At $x = b$ , the tangent line is horizontal so $f^{'} (b) = 0$ , and $f$ is differentiable.

At $x = c$ , the function has a jump discontinuity, so it is not differentiable at the point.

At $x = d$ , the vertical tangent line has a slope that is undefined, so $f^{'} (d)$ does not exist and is not differentiable at that point.

Example 3: Finding unknown variables

Consider the function $f (x)$ defined as:

$f (x) = ⎩ ⎨ ⎧ 3 x + 2 a x^{2} - b x if x < 1 if x \geq 1$

Find the values of $a$ and $b$ such that $f (x)$ is differentiable for all values of $x$ .

(spoiler)

Answer: $a = - 2, b = - 7$

If a piecewise function is differentiable, then it must be continuous.

For $f (x)$ to be continuous at $x = 1$ , both $f (1)$ and $x \to 1 lim f (x)$ must exist and be equal.

First,

$f (1) = a (1)^{2} - b (1) = a - b$

Next, compute the one-sided limits:

Left:

$x \to 1^{-} lim (3 x + 2) = 5$

Right:

$x \to 1^{+} lim (a x^{2} - b x) = a - b$

So continuity requires $a - b = 5$ .

Since there are two unknowns, we need a second equation. Differentiate each piece (treating $a$ and $b$ as constants):

$f^{'} (x) = ⎩ ⎨ ⎧ 3 2 a x - b if x < 1 if x > 1$

For differentiability at $x = 1$ , the one-sided derivatives must match:

$3 = 2 a (1) - b$

Now solve the system:

$a - b 2 a - b = 5 = 3$

which gives

$a = - 2 b = - 7$

Graph

$f (x) = ⎩ ⎨ ⎧ 3 x + 2 2 x^{2} + 7 x if x < 1 if x \geq 1$

in Desmos to confirm that the curve is unbroken and also smooth.

Achievable AP Calculus AB

2. Derivative basics

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.