2.6 Differentiability & continuity

Achievable AP Calculus AB

2. Derivative basics

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Differentiability & continuity

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What you’ll learn

Non-differentiability: Identify where a function fails to have a derivative using graphs or equations.
Differentiability vs. continuity: Apply the principle that differentiability implies continuity, but continuity does not guarantee 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 $x = a$ , meaning $f^{'} (a)$ does not exist:

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. 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 $f$ is not continuous at $x = a$ , then it’s not differentiable at that point $(not C \Rightarrow not D)$ .

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)

Discontinuous at $x = - 1$ (vertical asymptote), so $a$ is not differentiable 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 at $x = 1$ is yet to be determined. Check the one-sided derivatives:

Left: $x \to 1^{-} lim b^{'} (x) = - 1$
Right: $x \to 1^{+} lim b^{'} (x) = 1$

Since the two are unequal, $b^{'} (1)$ does not exist. Graphically, this shows up as a sharp corner at $x = 1$ .

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

(spoiler)

$c (x)$ is continuous with derivative

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

$c^{'} (0)$ is undefined (denominator equals $0$ ).

Notice the one-sided behavior:

Left: $x \to 0^{-} lim c^{'} (x) = - \infty$
Right: $x \to 0^{+} lim c^{'} (x) = + \infty$

Since the one-sided derivatives approach opposite infinities, the graph has a cusp at $x = 0$ .

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

(spoiler)

$d (x)$ is continuous with derivative

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

$d^{'} (0)$ is undefined (denominator equals $0$ ).

As $x$ approaches $0$ from either side, $d^{'} (x)$ approaches the same sign of $\infty$ (positive), meaning $d (x)$ has a vertical tangent line of $x = 0$ .

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 $f (x)$ differentiable?

Solution

(spoiler)

Answer: $x = b$ only

A function is differentiable only where it is continuous and has a finite slope.

$a$ : Corner (different one-sided derivatives) $\Rightarrow$ not differentiable.
$b$ : Horizontal tangent ( $f^{'} (b) = 0$ ) $\Rightarrow$ differentiable.
$c$ : Jump discontinuity $\Rightarrow$ not differentiable.
$d$ : Vertical tangent ( $f^{'} (d)$ doesn’t exist) $\Rightarrow$ not differentiable.

Example 3: Finding unknown variables

Let $f (x)$ be the piecewise function 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: $a = - 2, b = - 7$

For a piecewise function to be differentiable at $x = 1$ , it must be:

Continuous at $x = 1$
Have equal left and right derivatives at $x = 1$ .

1. Continuity:

This requires

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

Function value:

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

Left-hand limit:

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

Right-hand limit:

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

So continuity requires

$a - b = 5$

This provides one equation involving $a$ and $b$ . To solve for both variables, we now use differentiability at $x = 1$ .

2. Differentiability:

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.

Left-hand derivative: $x \to 1^{-} lim f^{'} (x) = 3$
Right-hand derivative: $x \to 1^{+} lim f^{'} (x) = 2 a - b$

Set equal:

$3 = 2 a - b$

3. Solve the system:

$a - b 2 a - b = 5 = 3 (1) (2)$

Solving the system of equations gives

$a = - 2 b = - 7$

as the values for $f$ to be differentiable.

Graph check (optional):

Graphing

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

in Desmos confirms the function is continuous and smooth at $x = 1$ .

Next | 3.1 Chain rule

Differentiability & continuity

What you’ll learn

Non-differentiability: Identify where a function fails to have a derivative using graphs or equations.
Differentiability vs. continuity: Apply the principle that differentiability implies continuity, but continuity does not guarantee 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 $x = a$ , meaning $f^{'} (a)$ does not exist:

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. 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 $f$ is not continuous at $x = a$ , then it’s not differentiable at that point $(not C \Rightarrow not D)$ .

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)

Discontinuous at $x = - 1$ (vertical asymptote), so $a$ is not differentiable 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 at $x = 1$ is yet to be determined. Check the one-sided derivatives:

Left: $x \to 1^{-} lim b^{'} (x) = - 1$
Right: $x \to 1^{+} lim b^{'} (x) = 1$

Since the two are unequal, $b^{'} (1)$ does not exist. Graphically, this shows up as a sharp corner at $x = 1$ .

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

(spoiler)

$c (x)$ is continuous with derivative

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

$c^{'} (0)$ is undefined (denominator equals $0$ ).

Notice the one-sided behavior:

Left: $x \to 0^{-} lim c^{'} (x) = - \infty$
Right: $x \to 0^{+} lim c^{'} (x) = + \infty$

Since the one-sided derivatives approach opposite infinities, the graph has a cusp at $x = 0$ .

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

(spoiler)

$d (x)$ is continuous with derivative

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

$d^{'} (0)$ is undefined (denominator equals $0$ ).

As $x$ approaches $0$ from either side, $d^{'} (x)$ approaches the same sign of $\infty$ (positive), meaning $d (x)$ has a vertical tangent line of $x = 0$ .

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 $f (x)$ differentiable?

Solution

(spoiler)

Answer: $x = b$ only

A function is differentiable only where it is continuous and has a finite slope.

$a$ : Corner (different one-sided derivatives) $\Rightarrow$ not differentiable.
$b$ : Horizontal tangent ( $f^{'} (b) = 0$ ) $\Rightarrow$ differentiable.
$c$ : Jump discontinuity $\Rightarrow$ not differentiable.
$d$ : Vertical tangent ( $f^{'} (d)$ doesn’t exist) $\Rightarrow$ not differentiable.

Example 3: Finding unknown variables

Let $f (x)$ be the piecewise function 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: $a = - 2, b = - 7$

For a piecewise function to be differentiable at $x = 1$ , it must be:

Continuous at $x = 1$
Have equal left and right derivatives at $x = 1$ .

1. Continuity:

This requires

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

Function value:

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

Left-hand limit:

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

Right-hand limit:

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

So continuity requires

$a - b = 5$

This provides one equation involving $a$ and $b$ . To solve for both variables, we now use differentiability at $x = 1$ .

2. Differentiability:

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.

Left-hand derivative: $x \to 1^{-} lim f^{'} (x) = 3$
Right-hand derivative: $x \to 1^{+} lim f^{'} (x) = 2 a - b$

Set equal:

$3 = 2 a - b$

3. Solve the system:

$a - b 2 a - b = 5 = 3 (1) (2)$

Solving the system of equations gives

$a = - 2 b = - 7$

as the values for $f$ to be differentiable.

Graph check (optional):

Graphing

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

in Desmos confirms the function is continuous and smooth at $x = 1$ .

Achievable AP Calculus AB

2. Derivative basics

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

Differentiability & continuity

7 min read

Font

Discuss

Feedback

What you’ll learn

Non-differentiability: Identify where a function fails to have a derivative using graphs or equations.
Differentiability vs. continuity: Apply the principle that differentiability implies continuity, but continuity does not guarantee 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 $x = a$ , meaning $f^{'} (a)$ does not exist:

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. 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 $f$ is not continuous at $x = a$ , then it’s not differentiable at that point $(not C \Rightarrow not D)$ .

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)

Discontinuous at $x = - 1$ (vertical asymptote), so $a$ is not differentiable 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 at $x = 1$ is yet to be determined. Check the one-sided derivatives:

Left: $x \to 1^{-} lim b^{'} (x) = - 1$
Right: $x \to 1^{+} lim b^{'} (x) = 1$

Since the two are unequal, $b^{'} (1)$ does not exist. Graphically, this shows up as a sharp corner at $x = 1$ .

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

(spoiler)

$c (x)$ is continuous with derivative

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

$c^{'} (0)$ is undefined (denominator equals $0$ ).

Notice the one-sided behavior:

Left: $x \to 0^{-} lim c^{'} (x) = - \infty$
Right: $x \to 0^{+} lim c^{'} (x) = + \infty$

Since the one-sided derivatives approach opposite infinities, the graph has a cusp at $x = 0$ .

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

(spoiler)

$d (x)$ is continuous with derivative

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

$d^{'} (0)$ is undefined (denominator equals $0$ ).

As $x$ approaches $0$ from either side, $d^{'} (x)$ approaches the same sign of $\infty$ (positive), meaning $d (x)$ has a vertical tangent line of $x = 0$ .

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 $f (x)$ differentiable?

Solution

(spoiler)

Answer: $x = b$ only

A function is differentiable only where it is continuous and has a finite slope.

$a$ : Corner (different one-sided derivatives) $\Rightarrow$ not differentiable.
$b$ : Horizontal tangent ( $f^{'} (b) = 0$ ) $\Rightarrow$ differentiable.
$c$ : Jump discontinuity $\Rightarrow$ not differentiable.
$d$ : Vertical tangent ( $f^{'} (d)$ doesn’t exist) $\Rightarrow$ not differentiable.

Example 3: Finding unknown variables

Let $f (x)$ be the piecewise function 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: $a = - 2, b = - 7$

For a piecewise function to be differentiable at $x = 1$ , it must be:

Continuous at $x = 1$
Have equal left and right derivatives at $x = 1$ .

1. Continuity:

This requires

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

Function value:

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

Left-hand limit:

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

Right-hand limit:

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

So continuity requires

$a - b = 5$

This provides one equation involving $a$ and $b$ . To solve for both variables, we now use differentiability at $x = 1$ .

2. Differentiability:

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.

Left-hand derivative: $x \to 1^{-} lim f^{'} (x) = 3$
Right-hand derivative: $x \to 1^{+} lim f^{'} (x) = 2 a - b$

Set equal:

$3 = 2 a - b$

3. Solve the system:

$a - b 2 a - b = 5 = 3 (1) (2)$

Solving the system of equations gives

$a = - 2 b = - 7$

as the values for $f$ to be differentiable.

Graph check (optional):

Graphing

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

in Desmos confirms the function is continuous and smooth at $x = 1$ .

Next | 3.1 Chain rule

Differentiability & continuity

What you’ll learn

Non-differentiability: Identify where a function fails to have a derivative using graphs or equations.
Differentiability vs. continuity: Apply the principle that differentiability implies continuity, but continuity does not guarantee 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 $x = a$ , meaning $f^{'} (a)$ does not exist:

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. 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 $f$ is not continuous at $x = a$ , then it’s not differentiable at that point $(not C \Rightarrow not D)$ .

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)

Discontinuous at $x = - 1$ (vertical asymptote), so $a$ is not differentiable 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 at $x = 1$ is yet to be determined. Check the one-sided derivatives:

Left: $x \to 1^{-} lim b^{'} (x) = - 1$
Right: $x \to 1^{+} lim b^{'} (x) = 1$

Since the two are unequal, $b^{'} (1)$ does not exist. Graphically, this shows up as a sharp corner at $x = 1$ .

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

(spoiler)

$c (x)$ is continuous with derivative

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

$c^{'} (0)$ is undefined (denominator equals $0$ ).

Notice the one-sided behavior:

Left: $x \to 0^{-} lim c^{'} (x) = - \infty$
Right: $x \to 0^{+} lim c^{'} (x) = + \infty$

Since the one-sided derivatives approach opposite infinities, the graph has a cusp at $x = 0$ .

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

(spoiler)

$d (x)$ is continuous with derivative

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

$d^{'} (0)$ is undefined (denominator equals $0$ ).

As $x$ approaches $0$ from either side, $d^{'} (x)$ approaches the same sign of $\infty$ (positive), meaning $d (x)$ has a vertical tangent line of $x = 0$ .

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 $f (x)$ differentiable?

Solution

(spoiler)

Answer: $x = b$ only

A function is differentiable only where it is continuous and has a finite slope.

$a$ : Corner (different one-sided derivatives) $\Rightarrow$ not differentiable.
$b$ : Horizontal tangent ( $f^{'} (b) = 0$ ) $\Rightarrow$ differentiable.
$c$ : Jump discontinuity $\Rightarrow$ not differentiable.
$d$ : Vertical tangent ( $f^{'} (d)$ doesn’t exist) $\Rightarrow$ not differentiable.

Example 3: Finding unknown variables

Let $f (x)$ be the piecewise function 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: $a = - 2, b = - 7$

For a piecewise function to be differentiable at $x = 1$ , it must be:

Continuous at $x = 1$
Have equal left and right derivatives at $x = 1$ .

1. Continuity:

This requires

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

Function value:

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

Left-hand limit:

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

Right-hand limit:

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

So continuity requires

$a - b = 5$

This provides one equation involving $a$ and $b$ . To solve for both variables, we now use differentiability at $x = 1$ .

2. Differentiability:

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.

Left-hand derivative: $x \to 1^{-} lim f^{'} (x) = 3$
Right-hand derivative: $x \to 1^{+} lim f^{'} (x) = 2 a - b$

Set equal:

$3 = 2 a - b$

3. Solve the system:

$a - b 2 a - b = 5 = 3 (1) (2)$

Solving the system of equations gives

$a = - 2 b = - 7$

as the values for $f$ to be differentiable.

Graph check (optional):

Graphing

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

in Desmos confirms the function is continuous and smooth at $x = 1$ .