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 said to be differentiable at $x = a$ if the 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 there.

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

2. Sharp corners

Where the function suddenly changes direction and the derivatives from each side exist but do not match.

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

3. Cusps

Similar to corners, but either one or both one-sided derivatives approach opposite infinities.

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

4. Vertical tangents

When the tangent line becomes a vertical line as the derivatives from either side both approach the same sign of infinity.

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

In the last three cases, each 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: Always check if $f (x)$ is continuous by verifying the conditions stated in Section 1.6: Continuity. If not continuous, then not differentiable.
Differentiate: Find $f^{'} (x)$ and check for any points where it is undefined.
If necessary (mostly for piecewise functions): The one-sided derivatives at $x = a$ must match and be a finite value $L$ i.e.

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

Examples

1. 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 $x = 0$ and is not continuous, so its derivative automatically does not exist there.

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

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

Because the derivative approaching from the left doesn’t match the one approaching the right, this results in a sharp corner at $x = 0$ in the original function, meaning there’s no value for the derivative there.

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

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

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

The derivative function is undefined when the denominator is 0, or when $x = 0$ .

Notice also that the derivative function approaches $- \infty$ when $x \to 0^{-}$ , while it approaches $\infty$ as $x \to 0^{+}$ . The one-sided derivatives approach opposite infinities which on the original function appears as a cusp at $x = 0$ .

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

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

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

The derivative is undefined when $x = 0$ .

Because $x$ in the denominator is squared, the derivative function approaches the same sign of $+ \infty$ on both sides of $x = 0$ , which appears as a vertical tangent on the original function’s graph.

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 satisfy the conditions for continuity at $x = 1$ , $f (1)$ and $x \to 1 lim f (x)$ must both exist and match.

First,

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

Next, one-sided limits:

Left:

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

Right:

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

They must agree, so $a - b = 5$ .

Having two unknown variables means we need two equations to solve for both. Taking the derivative of each piece (noting that $a$ and $b$ are simply constants):

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

To be differentiable, the limits of both must agree at $x = 1$ :

$3 = 2 a (1) - b$

Solving the system of equations

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

results in

$a = - 2 b = - 7$

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, let’s check for continuity by rewriting the absolute value function in piecewise form.

The breakpoints are when $x^{2} - x = 0$ . Factoring,

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

1. For $x < 0$ :
A point like $x = - 1$ results in a positive value, so the piece defined there is $+ (x^{2} - x)$ .

2. For $0 \leq x < 1$ :
A point like $x = 0.5$ results in a negative value, so the piece defined there is $- (x^{2} - x)$

3. For $x \geq 1$ :
A point like $x = 2$ results in a positive value, so the piece defined there is $+ (x^{2} - x)$

Rewritten in piecewise form,

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

Checking if the function is continuous at the breakpoint $x = 0$ ,

$f (0) = 0$

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

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

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

Checking if the function is continuous at the breakpoint $x = 1$ ,

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

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

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

Therefore the function is continuous.

Next, take the derivative of $f (x)$ :

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

Then at point of interest $x = 0$ :

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

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

Because the one-sided derivatives do not match, this function is not differentiable at $x = 0$ .

At the point of interest $x = 1$ :

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

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

So $f (x)$ is also not differentiable at $x = 1$ .

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

Key points

Differentiability implies continuity, but not vice versa.
A function that has a discontinuity, a sharp corner, a cusp, or a vertical tangent will not be differentiable there.
Always consider if a function is continuous before checking for differentiability.

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 said to be differentiable at $x = a$ if the 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 there.

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

2. Sharp corners

Where the function suddenly changes direction and the derivatives from each side exist but do not match.

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

3. Cusps

Similar to corners, but either one or both one-sided derivatives approach opposite infinities.

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

4. Vertical tangents

When the tangent line becomes a vertical line as the derivatives from either side both approach the same sign of infinity.

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

In the last three cases, each 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: Always check if $f (x)$ is continuous by verifying the conditions stated in Section 1.6: Continuity. If not continuous, then not differentiable.
Differentiate: Find $f^{'} (x)$ and check for any points where it is undefined.
If necessary (mostly for piecewise functions): The one-sided derivatives at $x = a$ must match and be a finite value $L$ i.e.

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

Examples

1. 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 $x = 0$ and is not continuous, so its derivative automatically does not exist there.

b) We already know that $b (x) = ∣ x ∣$ 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

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

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

The derivative function is undefined when the denominator is 0, or when $x = 0$ .

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

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

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

The derivative is undefined when $x = 0$ .

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 satisfy the conditions for continuity at $x = 1$ , $f (1)$ and $x \to 1 lim f (x)$ must both exist and match.

First,

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

Next, one-sided limits:

Left:

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

Right:

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

They must agree, so $a - b = 5$ .

Having two unknown variables means we need two equations to solve for both. Taking the derivative of each piece (noting that $a$ and $b$ are simply constants):

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

To be differentiable, the limits of both must agree at $x = 1$ :

$3 = 2 a (1) - b$

Solving the system of equations

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

results in

$a = - 2 b = - 7$

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, let’s check for continuity by rewriting the absolute value function in piecewise form.

The breakpoints are when $x^{2} - x = 0$ . Factoring,

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

1. For $x < 0$ :
A point like $x = - 1$ results in a positive value, so the piece defined there is $+ (x^{2} - x)$ .

2. For $0 \leq x < 1$ :
A point like $x = 0.5$ results in a negative value, so the piece defined there is $- (x^{2} - x)$

3. For $x \geq 1$ :
A point like $x = 2$ results in a positive value, so the piece defined there is $+ (x^{2} - x)$

Rewritten in piecewise form,

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

Checking if the function is continuous at the breakpoint $x = 0$ ,

$f (0) = 0$

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

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

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

Checking if the function is continuous at the breakpoint $x = 1$ ,

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

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

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

Therefore the function is continuous.

Next, take the derivative of $f (x)$ :

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

Then at point of interest $x = 0$ :

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

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

Because the one-sided derivatives do not match, this function is not differentiable at $x = 0$ .

At the point of interest $x = 1$ :

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

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

So $f (x)$ is also not differentiable at $x = 1$ .

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

Key points

Differentiability implies continuity, but not vice versa.
A function that has a discontinuity, a sharp corner, a cusp, or a vertical tangent will not be differentiable there.
Always consider if a function is continuous before checking for differentiability.

Differentiability & continuity

What you’ll learn:

What is differentiability?

Checking for differentiability

Examples

Solutions

Solution

Solution

Differentiability & continuity

What you’ll learn:

What is differentiability?

Checking for differentiability

Examples

Solutions

Solution

Solution