2.6 Differentiability & continuity

Achievable AP Calculus AB

2. Derivative basics

Our AP Calculus AB course is currently in development and is a work-in-progress.

Differentiability & continuity

7 min read

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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 $x = a$ :

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 there.

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

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. $f (x) = ∣ x ∣$

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. $f (x) = x^{2/3}$

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:

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

Consequently, the contrapositive is always true:

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

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 there.
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

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

Examples

Consider the functions:

a) $a (x) = 1/ x$
b) $b (x) = ∣ x ∣$
c) $c (x) = x^{2/3}$
d) $d (x) = x^{1/3}$

For what $x$ -value(s) is each non-differentiable and why?

Solutions

a) $\frac{1}{x}$ has a vertical asymptote at $x = 0$ , so it is not continuous there. Since differentiability implies continuity, the derivative does not exist at $x = 0$ .

b) $b (x) = ∣ x ∣$ is continuous everywhere. Its derivative is the piecewise function:

$b^{'} (x) = ⎩ ⎨ ⎧ - 1 1 if x < 0 if x > 0$

As $x \to 0^{-}$ , the derivative approaches $- 1$ , and as $x \to 0^{+}$ , it approaches $1$ . Because these one-sided derivatives don’t match, $b^{'} (0)$ does not exist. On the graph, this shows up as a sharp corner at $x = 0$ .

c) The derivative of $c (x) = x^{2/3}$ is

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

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

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 there.

d) The derivative of $d (x) = x^{1/3}$ is

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

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

This derivative is undefined at $x = 0$ .

Because $x$ is squared in the denominator, $3 x^{2}$ is positive for $x \neq = 0$ . That means $d^{'} (x) \to + \infty$ from both sides as $x \to 0$ . On the graph, this appears as a vertical tangent at $x = 0$ , so the function is not differentiable there.

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 (everywhere).

Solution

(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:

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

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:

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

This 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.

Identify all the points where $f (x) = ∣ x^{2} - x ∣$ is not differentiable.

Solution

(spoiler)

First, check where the expression inside the absolute value changes sign by solving $x^{2} - x = 0$ .

Factoring,

$x (x - 1) = 0 x (x - 1) = 0 x = 0, 1$

These are the breakpoints for a piecewise definition.

For $x < 0$ :
A test value like $x = - 1$ makes $x^{2} - x$ positive, so $∣ x^{2} - x ∣ = + (x^{2} - x)$ .
For $0 \leq x < 1$ :
A test value like $x = 0.5$ makes $x^{2} - x$ negative, so $∣ x^{2} - x ∣ = - (x^{2} - x)$ .
For $x \geq 1$ :
A test value like $x = 2$ makes $x^{2} - x$ positive, so $∣ x^{2} - x ∣ = + (x^{2} - x)$ .

Rewritten in piecewise form,

$f (x) = ⎩ ⎨ ⎧ x^{2} - x - (x^{2} - x) x^{2} - x if x < 0 if 0 \leq x < 1 if x \geq 1$

Now check continuity at each breakpoint.

At $x = 0$ :

$f (0) = 0$

$x \to 0^{-} lim (x^{2} - x) = 0$

$x \to 0^{+} lim (- x^{2} + x) = 0$

So the function is continuous at $x = 0$ .

At $x = 1$ :

$f (1) = (1)^{2} - 1 = 0$

$x \to 1^{-} lim (- x^{2} + x) = 0$

$x \to 1^{+} lim (x^{2} - x) = 0$

So the function is also continuous at $x = 1$ .

Next, take the derivative on each interval:

$f^{'} (x) = ⎩ ⎨ ⎧ 2 x - 1 - 2 x + 1 2 x - 1 if x < 0 if 0 < x < 1 if x > 1$

Now compare one-sided derivatives at the breakpoints.

At $x = 0$ :

$x \to 0^{-} lim (2 x - 1) = - 1$

$x \to 0^{+} lim (- 2 x + 1) = 1$

The one-sided derivatives don’t match, so $f (x)$ is not differentiable at $x = 0$ .

At $x = 1$ :

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

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

The one-sided derivatives don’t match, so $f (x)$ is not differentiable at $x = 1$ .

Confirm visually with the graph of $f (x)$ to see the sharp corners at $x = 0$ and $x = 1$ .

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 $x = a$ :

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 there.

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

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. $f (x) = ∣ x ∣$

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. $f (x) = x^{2/3}$

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:

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

Consequently, the contrapositive is always true:

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

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 there.
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

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

Examples

Consider the functions:

a) $a (x) = 1/ x$
b) $b (x) = ∣ x ∣$
c) $c (x) = x^{2/3}$
d) $d (x) = x^{1/3}$

For what $x$ -value(s) is each non-differentiable and why?

Solutions

a) $\frac{1}{x}$ has a vertical asymptote at $x = 0$ , so it is not continuous there. Since differentiability implies continuity, the derivative does not exist at $x = 0$ .

b) $b (x) = ∣ x ∣$ is continuous everywhere. Its derivative is the piecewise function:

$b^{'} (x) = ⎩ ⎨ ⎧ - 1 1 if x < 0 if x > 0$

c) The derivative of $c (x) = x^{2/3}$ is

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

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

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 there.

d) The derivative of $d (x) = x^{1/3}$ is

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

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

This derivative is undefined at $x = 0$ .

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 (everywhere).

Solution

(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:

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

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:

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

This 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.

Identify all the points where $f (x) = ∣ x^{2} - x ∣$ is not differentiable.

Solution

(spoiler)

First, check where the expression inside the absolute value changes sign by solving $x^{2} - x = 0$ .

Factoring,

$x (x - 1) = 0 x (x - 1) = 0 x = 0, 1$

These are the breakpoints for a piecewise definition.

For $x < 0$ :
A test value like $x = - 1$ makes $x^{2} - x$ positive, so $∣ x^{2} - x ∣ = + (x^{2} - x)$ .
For $0 \leq x < 1$ :
A test value like $x = 0.5$ makes $x^{2} - x$ negative, so $∣ x^{2} - x ∣ = - (x^{2} - x)$ .
For $x \geq 1$ :
A test value like $x = 2$ makes $x^{2} - x$ positive, so $∣ x^{2} - x ∣ = + (x^{2} - x)$ .

Rewritten in piecewise form,

$f (x) = ⎩ ⎨ ⎧ x^{2} - x - (x^{2} - x) x^{2} - x if x < 0 if 0 \leq x < 1 if x \geq 1$

Now check continuity at each breakpoint.

At $x = 0$ :

$f (0) = 0$

$x \to 0^{-} lim (x^{2} - x) = 0$

$x \to 0^{+} lim (- x^{2} + x) = 0$

So the function is continuous at $x = 0$ .

At $x = 1$ :

$f (1) = (1)^{2} - 1 = 0$

$x \to 1^{-} lim (- x^{2} + x) = 0$

$x \to 1^{+} lim (x^{2} - x) = 0$

So the function is also continuous at $x = 1$ .

Next, take the derivative on each interval:

$f^{'} (x) = ⎩ ⎨ ⎧ 2 x - 1 - 2 x + 1 2 x - 1 if x < 0 if 0 < x < 1 if x > 1$

Now compare one-sided derivatives at the breakpoints.

At $x = 0$ :

$x \to 0^{-} lim (2 x - 1) = - 1$

$x \to 0^{+} lim (- 2 x + 1) = 1$

The one-sided derivatives don’t match, so $f (x)$ is not differentiable at $x = 0$ .

At $x = 1$ :

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

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

The one-sided derivatives don’t match, so $f (x)$ is not differentiable at $x = 1$ .

Confirm visually with the graph of $f (x)$ to see the sharp corners at $x = 0$ and $x = 1$ .

Achievable AP Calculus AB

2. Derivative basics

Our AP Calculus AB course is currently in development and is a work-in-progress.

Differentiability & continuity

7 min read

Font

Discuss

Feedback

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 $x = a$ :

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 there.

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

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. $f (x) = ∣ x ∣$

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. $f (x) = x^{2/3}$

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:

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

Consequently, the contrapositive is always true:

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

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 there.
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

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

Examples

Consider the functions:

a) $a (x) = 1/ x$
b) $b (x) = ∣ x ∣$
c) $c (x) = x^{2/3}$
d) $d (x) = x^{1/3}$

For what $x$ -value(s) is each non-differentiable and why?

Solutions

a) $\frac{1}{x}$ has a vertical asymptote at $x = 0$ , so it is not continuous there. Since differentiability implies continuity, the derivative does not exist at $x = 0$ .

b) $b (x) = ∣ x ∣$ is continuous everywhere. Its derivative is the piecewise function:

$b^{'} (x) = ⎩ ⎨ ⎧ - 1 1 if x < 0 if x > 0$

c) The derivative of $c (x) = x^{2/3}$ is

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

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

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 there.

d) The derivative of $d (x) = x^{1/3}$ is

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

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

This derivative is undefined at $x = 0$ .

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 (everywhere).

Solution

(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:

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

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:

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

This 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.

Identify all the points where $f (x) = ∣ x^{2} - x ∣$ is not differentiable.

Solution

(spoiler)

First, check where the expression inside the absolute value changes sign by solving $x^{2} - x = 0$ .

Factoring,

$x (x - 1) = 0 x (x - 1) = 0 x = 0, 1$

These are the breakpoints for a piecewise definition.

For $x < 0$ :
A test value like $x = - 1$ makes $x^{2} - x$ positive, so $∣ x^{2} - x ∣ = + (x^{2} - x)$ .
For $0 \leq x < 1$ :
A test value like $x = 0.5$ makes $x^{2} - x$ negative, so $∣ x^{2} - x ∣ = - (x^{2} - x)$ .
For $x \geq 1$ :
A test value like $x = 2$ makes $x^{2} - x$ positive, so $∣ x^{2} - x ∣ = + (x^{2} - x)$ .

Rewritten in piecewise form,

$f (x) = ⎩ ⎨ ⎧ x^{2} - x - (x^{2} - x) x^{2} - x if x < 0 if 0 \leq x < 1 if x \geq 1$

Now check continuity at each breakpoint.

At $x = 0$ :

$f (0) = 0$

$x \to 0^{-} lim (x^{2} - x) = 0$

$x \to 0^{+} lim (- x^{2} + x) = 0$

So the function is continuous at $x = 0$ .

At $x = 1$ :

$f (1) = (1)^{2} - 1 = 0$

$x \to 1^{-} lim (- x^{2} + x) = 0$

$x \to 1^{+} lim (x^{2} - x) = 0$

So the function is also continuous at $x = 1$ .

Next, take the derivative on each interval:

$f^{'} (x) = ⎩ ⎨ ⎧ 2 x - 1 - 2 x + 1 2 x - 1 if x < 0 if 0 < x < 1 if x > 1$

Now compare one-sided derivatives at the breakpoints.

At $x = 0$ :

$x \to 0^{-} lim (2 x - 1) = - 1$

$x \to 0^{+} lim (- 2 x + 1) = 1$

The one-sided derivatives don’t match, so $f (x)$ is not differentiable at $x = 0$ .

At $x = 1$ :

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

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

The one-sided derivatives don’t match, so $f (x)$ is not differentiable at $x = 1$ .

Confirm visually with the graph of $f (x)$ to see the sharp corners at $x = 0$ and $x = 1$ .

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 $x = a$ :

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 there.

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

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. $f (x) = ∣ x ∣$

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. $f (x) = x^{2/3}$

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:

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

Consequently, the contrapositive is always true:

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

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 there.
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

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

Examples

Consider the functions:

a) $a (x) = 1/ x$
b) $b (x) = ∣ x ∣$
c) $c (x) = x^{2/3}$
d) $d (x) = x^{1/3}$

For what $x$ -value(s) is each non-differentiable and why?

Solutions

a) $\frac{1}{x}$ has a vertical asymptote at $x = 0$ , so it is not continuous there. Since differentiability implies continuity, the derivative does not exist at $x = 0$ .

b) $b (x) = ∣ x ∣$ is continuous everywhere. Its derivative is the piecewise function:

$b^{'} (x) = ⎩ ⎨ ⎧ - 1 1 if x < 0 if x > 0$

c) The derivative of $c (x) = x^{2/3}$ is

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

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

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 there.

d) The derivative of $d (x) = x^{1/3}$ is

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

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

This derivative is undefined at $x = 0$ .

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 (everywhere).

Solution

(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:

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

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:

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

This 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.

Identify all the points where $f (x) = ∣ x^{2} - x ∣$ is not differentiable.

Solution

(spoiler)

First, check where the expression inside the absolute value changes sign by solving $x^{2} - x = 0$ .

Factoring,

$x (x - 1) = 0 x (x - 1) = 0 x = 0, 1$

These are the breakpoints for a piecewise definition.

For $x < 0$ :
A test value like $x = - 1$ makes $x^{2} - x$ positive, so $∣ x^{2} - x ∣ = + (x^{2} - x)$ .
For $0 \leq x < 1$ :
A test value like $x = 0.5$ makes $x^{2} - x$ negative, so $∣ x^{2} - x ∣ = - (x^{2} - x)$ .
For $x \geq 1$ :
A test value like $x = 2$ makes $x^{2} - x$ positive, so $∣ x^{2} - x ∣ = + (x^{2} - x)$ .

Rewritten in piecewise form,

$f (x) = ⎩ ⎨ ⎧ x^{2} - x - (x^{2} - x) x^{2} - x if x < 0 if 0 \leq x < 1 if x \geq 1$

Now check continuity at each breakpoint.

At $x = 0$ :

$f (0) = 0$

$x \to 0^{-} lim (x^{2} - x) = 0$

$x \to 0^{+} lim (- x^{2} + x) = 0$

So the function is continuous at $x = 0$ .

At $x = 1$ :

$f (1) = (1)^{2} - 1 = 0$

$x \to 1^{-} lim (- x^{2} + x) = 0$

$x \to 1^{+} lim (x^{2} - x) = 0$

So the function is also continuous at $x = 1$ .

Next, take the derivative on each interval:

$f^{'} (x) = ⎩ ⎨ ⎧ 2 x - 1 - 2 x + 1 2 x - 1 if x < 0 if 0 < x < 1 if x > 1$

Now compare one-sided derivatives at the breakpoints.

At $x = 0$ :

$x \to 0^{-} lim (2 x - 1) = - 1$

$x \to 0^{+} lim (- 2 x + 1) = 1$

The one-sided derivatives don’t match, so $f (x)$ is not differentiable at $x = 0$ .

At $x = 1$ :

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

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

The one-sided derivatives don’t match, so $f (x)$ is not differentiable at $x = 1$ .

Confirm visually with the graph of $f (x)$ to see the sharp corners at $x = 0$ and $x = 1$ .