Does differentiable always mean continuous?

Yes. This is the main theorem of the lesson: if a function is differentiable at a point, it is automatically continuous there. The proof works by writing f(x)-f(a) as the difference quotient times (x-a), whose limit is f'(a)· 0=0.

Why doesn't continuity imply differentiability?

Because a continuous graph can still have a sharp point or an infinite slope. The classic example is |x| at x=0: it is continuous but has different left and right slopes, so no single derivative value exists. Continuity only rules out breaks, not sharp turns.

What are the three ways a continuous function can fail to be differentiable?

A corner (finite but unequal one-sided slopes, like |x|), a cusp (opposite infinite slopes, like x^2/3), and a vertical tangent (same infinite slope from both sides, like x^1/3). In all three the function stays continuous.

How do I check if a piecewise function is differentiable at the boundary?

Check two conditions. First, the pieces must give the same value at the boundary (continuity). Second, the derivatives of the two pieces must be equal at the boundary (matching slopes). Both must hold; matching values alone is not enough.

AP-CALCBC-2.4

U2.4 Connecting Differentiability and Continuity

Learn why differentiability implies continuity, why the converse fails, and how to spot corners, cusps, and vertical tangents on the AP Calculus BC exam.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U2.4 Connecting Differentiability and Continuity, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

You already know how to compute a derivative as a limit. Now comes one of the most tested logical relationships in all of calculus: if a function is differentiable at a point, it must be continuous there. This lesson proves that statement, then shows you the crucial subtlety — the reverse is not true. A function can be perfectly continuous yet fail to have a derivative at a point.

By the end you will be able to state and prove the theorem, use it to rule out differentiability, and instantly classify the three ways a continuous function goes non-differentiable: corners, cusps, and vertical tangents. The AP exam loves to test whether you understand the direction of this implication.

The Theorem and Its Proof

The central result is short: if $f$ is differentiable at $x=a$ , then $f$ is continuous at $x=a$ .

To prove continuity at

a

we must show

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

, which is equivalent to showing

\lim_{x\to a}\left[f(x)-f(a)\right]=0

. Here is the trick: multiply and divide by

(x-a)

\lim_{x\to a}\left[f(x)-f(a)\right]=\lim_{x\to a}\left[\frac{f(x)-f(a)}{x-a}\cdot(x-a)\right]

Because

f

is differentiable at

a

, the first factor has a finite limit equal to

f'(a)

. The second factor,

(x-a)

, approaches

0

. Using the product rule for limits:

\lim_{x\to a}\left[f(x)-f(a)\right]=f'(a)\cdot 0 = 0.

Therefore

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

, which is exactly the definition of continuity at

a

.

The key step is that differentiability guarantees the difference quotient converges to a finite number. That finiteness is what forces the product to collapse to zero. If the difference quotient blew up to infinity, this argument would fail — which foreshadows the vertical tangent failure mode. On the exam, you may be asked to reproduce this proof or to justify a single step, so understand why the

(x-a)

factor is inserted.

The Converse Fails: Continuity Does Not Imply Differentiability

The theorem is a one-way street. Continuity is a necessary condition for differentiability, but not a sufficient one. In logic terms, differentiable

\Rightarrow

continuous, but continuous

\not\Rightarrow

differentiable.

The contrapositive of the theorem is extremely useful: if $f$ is not continuous at $a$ , then $f$ is not differentiable at $a$ . So any jump, hole, or asymptote automatically kills differentiability. This gives you a fast elimination tool.

The famous counterexample to the converse is

f(x)=|x|

x=0

. It is continuous everywhere, yet its derivative does not exist at

0

because the left-hand slope is

-1

and the right-hand slope is

+1

. The limit of the difference quotient depends on direction, so no single value exists.

Statement	True?
Differentiable at $a$ $\Rightarrow$ continuous at $a$	Yes
Continuous at $a$ $\Rightarrow$ differentiable at $a$	No
Not continuous at $a$ $\Rightarrow$ not differentiable at $a$	Yes
Not differentiable at $a$ $\Rightarrow$ not continuous at $a$	No

A common misconception is treating the two properties as equivalent. They are not. Differentiability is the stronger condition: every differentiable function is continuous, but the reverse can break down at isolated points.

The Three Failure Modes

When a function is continuous but not differentiable at a point, it happens in one of three geometrically distinct ways. Knowing these lets you diagnose a graph instantly.

A corner occurs when the one-sided derivatives are both finite but unequal. The graph makes a sharp turn. Example:

f(x)=|x|

x=0

, with left slope

-1

and right slope

+1

.

A cusp occurs when the one-sided slopes approach

+\infty

and

-\infty

(or vice versa), pinching to a sharp point. Example:

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

x=0

. Here

f'(x)=\tfrac{2}{3}x^{-1/3}

, which tends to

-\infty

from the left and

+\infty

from the right.

A vertical tangent occurs when the slope approaches

+\infty

(or

-\infty

) from both sides. The tangent line is vertical, so it has no finite slope. Example:

f(x)=x^{1/3}

x=0

, where

f'(x)=\tfrac{1}{3}x^{-2/3}\to+\infty

on both sides.

Failure mode	Left slope	Right slope	Example at $x=0$
Corner	finite	finite, different	$\|x\|$
Cusp	$\pm\infty$	$\mp\infty$	$x^{2/3}$
Vertical tangent	$\pm\infty$	same $\pm\infty$	$x^{1/3}$

In all three cases the function is continuous — no break in the graph — but the derivative limit fails to be a single finite number.

How the Exam Tests This

AP questions on this topic take a few predictable forms. First, multiple-choice items may show a piecewise function or a graph and ask where the derivative fails to exist. Scan for discontinuities first (using the contrapositive), then look for corners, cusps, and vertical tangents.

Second, free-response and justification questions may require you to check differentiability of a piecewise function at a boundary. The standard method has two conditions that must both hold: the pieces must match in value (continuity), and the derivatives of the two pieces must match at the boundary (matching slopes). If a problem gives

f(x)

defined one way for

x\le a

and another for

x>a

with unknown constants, you typically set up two equations.

Third, conceptual items test the logic directly: 'If

f

is continuous at

x=3

, must

f

be differentiable at

x=3

?' The correct answer is no, and a full-credit justification names a counterexample or a failure mode.

A frequent trap: students conclude a function is differentiable just because it is continuous. Another trap is forgetting that matching values alone is not enough — you also need matching slopes. Always verify both conditions. When justifying, use precise language: state that the left- and right-hand derivatives exist and are equal to a finite common value, or explain exactly which condition fails.

Key terms

Differentiable at $a$ .: The limit $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ exists as a finite number; equivalently, the one-sided difference quotients agree on a single finite value.
Continuous at $a$ .: $\lim_{x\to a} f(x)=f(a)$ ; the function has no jump, hole, or asymptote at $a$ .
Converse.: Reversing an implication. The converse of 'differentiable $\Rightarrow$ continuous' is 'continuous $\Rightarrow$ differentiable,' which is false.
Contrapositive.: Logically equivalent restatement: 'not continuous $\Rightarrow$ not differentiable.' A powerful elimination tool for spotting non-differentiable points.
Corner.: A point where both one-sided derivatives are finite but unequal, producing a sharp turn (e.g. $|x|$ at $0$ ).
Cusp.: A point where the one-sided slopes head to $+\infty$ and $-\infty$ (opposite infinities), pinching to a sharp point (e.g. $x^{2/3}$ at $0$ ).
Vertical tangent.: A point where the slope approaches the same infinite value from both sides, giving a vertical tangent line and no finite derivative (e.g. $x^{1/3}$ at $0$ ).

Worked example

Let

f(x)=\begin{cases} x^2+1 & x\le 1 \\ ax+b & x>1 \end{cases}

. Find constants

a

and

b

so that

f

is differentiable at

x=1

Differentiability at

x=1

requires two things: continuity at

1

(values match) and matching derivatives at

1

(slopes match). We use both conditions.

First, continuity. The left piece gives

f(1)=1^2+1=2

. The right piece approaches

a(1)+b=a+b

x\to 1^+

. Setting them equal:

a+b=2.

Second, matching slopes. Differentiate each piece. For

x\le 1

f'(x)=2x

, so the left derivative at

1

2(1)=2

. For

x>1

f'(x)=a

. Setting the one-sided derivatives equal:

a=2.

Substitute

a=2

into

a+b=2

to get

2+b=2

, so

b=0

.

Therefore

a=2

and

b=0

. With these values the function is continuous (both pieces give

2

x=1

) and has a common slope of

2

from both sides, so it is differentiable at

x=1

.

Note the logic: had we only required continuity, infinitely many

(a,b)

pairs would work. The slope condition, guaranteed necessary by differentiability, pins down the unique answer.

Practice questions

The function

g

is continuous at

x=4

. Which of the following must be true?

$g$ is differentiable at $x=4$
$g'(4)$ exists and equals $0$
$\lim_{x\to 4} g(x)=g(4)$
$g$ has a corner at $x=4$

Answer: $\lim_{x\to 4} g(x)=g(4)$

Continuity at

4

means exactly that

\lim_{x\to 4} g(x)=g(4)

, so that choice is the definition and must be true. Continuity does not guarantee differentiability, so the first and second choices are not required. A corner is only one possible behavior, not a necessity, so the last choice is false.

Consider

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

. Determine whether

f

is continuous at

x=0

and whether it is differentiable there. Justify your answer and name the type of behavior at

x=0

Answer: $f$ is continuous at $0$ but not differentiable there; the graph has a cusp.

Continuity:

\lim_{x\to 0} x^{2/3}=0=f(0)

, so

f

is continuous at

0

. Differentiability:

f'(x)=\tfrac{2}{3}x^{-1/3}

. As

x\to 0^-

this tends to

-\infty

and as

x\to 0^+

it tends to

+\infty

. The one-sided slopes head to opposite infinities, so the derivative does not exist. Opposite infinite slopes pinching to a point is the definition of a cusp. This illustrates that continuity does not imply differentiability.

A function

h

has a jump discontinuity at

x=2

. Is

h

differentiable at

x=2

? Explain using a logical principle from this lesson.

Answer: No, $h$ is not differentiable at $x=2$ .

Use the contrapositive of the theorem 'differentiable

\Rightarrow

continuous.' The contrapositive states 'not continuous

\Rightarrow

not differentiable.' A jump discontinuity means

h

is not continuous at

x=2

, so

h

cannot be differentiable there. Differentiability requires continuity as a necessary condition, and that condition fails.

FAQ

Does differentiable always mean continuous?: Yes. This is the main theorem of the lesson: if a function is differentiable at a point, it is automatically continuous there. The proof works by writing $f(x)-f(a)$ as the difference quotient times $(x-a)$ , whose limit is $f'(a)\cdot 0=0$ .
Why doesn't continuity imply differentiability?: Because a continuous graph can still have a sharp point or an infinite slope. The classic example is $|x|$ at $x=0$ : it is continuous but has different left and right slopes, so no single derivative value exists. Continuity only rules out breaks, not sharp turns.
What are the three ways a continuous function can fail to be differentiable?: A corner (finite but unequal one-sided slopes, like $|x|$ ), a cusp (opposite infinite slopes, like $x^{2/3}$ ), and a vertical tangent (same infinite slope from both sides, like $x^{1/3}$ ). In all three the function stays continuous.
How do I check if a piecewise function is differentiable at the boundary?: Check two conditions. First, the pieces must give the same value at the boundary (continuity). Second, the derivatives of the two pieces must be equal at the boundary (matching slopes). Both must hold; matching values alone is not enough.

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