Connecting differentiability and continuity

AP Calculus BC· AP Calculus BC CED — Differentiation: Definition and Fundamental Properties· 14 min read

1. Theorem: Differentiability Implies Continuity★★☆☆☆⏱ 4 min

The core result of this topic is a formal logical implication: if a function $f$ is differentiable at $x = a$ , then $f$ must be continuous at $x = a$ . This can be proven directly using limit laws, and its contrapositive is one of the most useful tools for quickly testing differentiability on the AP exam.

🔬 Derivation

Goal:

Prove that differentiability at $x = a$ implies continuity at $x = a$

Starting from:

If $f$ is differentiable at $a$ , the derivative $f^{'} (a) = lim_{h \to 0} \frac{f ( a + h ) - f ( a )}{h}$ exists as a finite limit.

1
To confirm continuity, we need to show $lim_{h \to 0} [f (a + h) - f (a)] = 0$ , the definition of continuity at $x = a$ .
2
$\lim_{h \to 0} \left[f(a+h) - f(a)\right] = \lim_{h \to 0} \left( \frac{f(a+h) - f(a)}{h} \cdot h \right)$
3
Split using the product rule for limits, which applies because both limits exist:
4
$\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \cdot \lim_{h \to 0} h = f'(a) \cdot 0 = 0$

Result:

We have shown $lim_{h \to 0} [f (a + h) - f (a)] = 0$ , so $f$ is continuous at $x = a$ . The contrapositive of the theorem is: If $f$ is not continuous at $x = a$ , then $f$ is not differentiable at $x = a$ . This lets you stop testing for differentiability immediately if you find a discontinuity, saving time on the exam.

📐 Worked Example

Let $f (x) = {3 x - 2 3 x + 1 x < 2 x \geq 2$ . Is $f$ differentiable at $x = 2$ ? Justify your answer.

1
First check continuity at $x = 2$ by calculating the left-hand limit:
2
$\lim_{x \to 2^-} f(x) = 3(2) - 2 = 4$
3
Calculate the right-hand limit and function value at $x = 2$ :
4
$\lim_{x \to 2^+} f(x) = f(2) = 3(2) + 1 = 7$
5
Since the left-hand limit does not equal the function value, $f$ is discontinuous at $x = 2$ .
6
By the contrapositive of the differentiability-implies-continuity theorem, $f$ cannot be differentiable at $x = 2$ .

Exam tip:

Always check for discontinuity first when testing differentiability. If you find a discontinuity, you can stop immediately and write your conclusion, saving 1-2 minutes on exam day.

2. Continuity Does Not Imply Differentiability★★★☆☆⏱ 5 min

While differentiability guarantees continuity, the converse is not true: continuity is a necessary, but not sufficient, condition for differentiability. A function can be fully continuous at $x = a$ but still fail to be differentiable there. There are four common cases of continuous non-differentiable functions:

Corners: left-hand derivative $\neq =$ right-hand derivative
Cusps: one-sided derivatives approach opposite infinities
Vertical tangents: derivative approaches $\pm \infty$ (not finite)
Oscillating tangents: the limit of the difference quotient does not exist

On the AP exam, this concept is most commonly tested for piecewise functions that are continuous at their junction point: you must check if the left-hand derivative equals the right-hand derivative to confirm differentiability.

📐 Worked Example

Let $f (x) = {x^{2} + 3 7 x - 9 x \leq 3 x > 3$ . Is $f$ differentiable at $x = 3$ ? Justify your answer.

1
First confirm continuity at $x = 3$ :
2
$f(3) = 3^2 + 3 = 12, \quad \lim_{x \to 3^+} f(x) = 7(3) - 9 = 12$
3
So $f$ is continuous at $x = 3$ . Next calculate the left-hand derivative, which equals the derivative of the left piece evaluated at $x = 3$ :
4
$f'_-(3) = \frac{d}{dx}(x^2 + 3)\bigg|_{x=3} = 2(3) = 6$
5
Calculate the right-hand derivative similarly:
6
$f'_+(3) = \frac{d}{dx}(7x - 9)\bigg|_{x=3} = 7$
7
Since $f_{-}^{'} (3) = 6 \neq = 7 = f_{+}^{'} (3)$ , the two-sided derivative does not exist, so $f$ is continuous but not differentiable at $x = 3$ .

Exam tip:

Do not just compare derivatives of the pieces and skip the continuity check. If the function is discontinuous, equal derivatives on each side do not make it differentiable.

3. Identifying Non-Differentiable Points From Graphs★★☆☆☆⏱ 3 min

A common AP multiple-choice question gives you the graph of $f (x)$ and asks you to count how many non-differentiable points exist in a given interval. To solve these, check for any of the four types of non-differentiable points directly on the graph.

📐 Worked Example

The graph of $y = f (x)$ on the open interval $(- 4, 4)$ has a jump discontinuity at $x = - 2$ , a corner at $x = 0$ , a vertical tangent at $x = 2$ , and is smooth at all other points. How many non-differentiable points are in $(- 4, 4)$ ?

1
List all candidate points inside the open interval: $x = - 2$ , $x = 0$ , $x = 2$ .
2
$x = - 2$ is a discontinuity: by the contrapositive theorem, it is non-differentiable (count = 1).
3
$x = 0$ is a corner: it is continuous, but the left slope does not equal the right slope, so it is non-differentiable (count = 2).
4
$x = 2$ has a vertical tangent: the slope is infinite, so it is not a finite derivative, hence non-differentiable (count = 3).
5
There are no other candidate points, so the total number of non-differentiable points is 3.

Exam tip:

Don't forget to count vertical tangents as non-differentiable. Most students remember corners and discontinuities but miss vertical tangents on graph problems.

4. Finding Constants for Differentiable Piecewise Functions★★★★☆⏱ 6 min

A common free-response question asks you to find unknown constants in a piecewise function such that the function is both continuous and differentiable at the junction point. Always follow the order: find constants for continuity first, then solve for differentiability.

📐 Worked Example

Let $g (x) = {a x^{3} + b x 4 x^{2} - 2 x \leq 1 x > 1$ . (a) Find all values of $a$ and $b$ such that $g (x)$ is continuous at $x = 1$ . (b) For continuous $g$ , when is $g$ also differentiable at $x = 1$ ? (c) Suppose $a = 3$ and $b = - 1$ . Is $g$ differentiable at $x = 1$ ? Justify.

1
Part (a): For continuity, the left-hand limit must equal the right-hand limit at $x = 1$ :
2
$\lim_{x \to 1^+} (4x^2 - 2) = 4(1) - 2 = 2, \quad \lim_{x \to 1^-} (ax^3 + bx) = a + b$
3
All pairs $a, b$ satisfying $a + b = 2$ make $g$ continuous at $x = 1$ .
4
Part (b): For differentiability, compare one-sided derivatives after confirming continuity:
5
$g'_+(1) = \frac{d}{dx}(4x^2 - 2)\bigg|_{x=1} = 8, \quad g'_-(1) = 3a + b$
6
Set derivatives equal, then substitute $b = 2 - a$ from part (a):
7
$3a + (2 - a) = 8 \implies 2a = 6 \implies a = 3, b = -1$
8
Part (c): For $a = 3, b = - 1$ , $a + b = 2$ so $g$ is continuous at $x = 1$ . Both one-sided derivatives equal 8, so $g$ is differentiable at $x = 1$ .

✓ Quick check

Test your understanding of the core implication:

Let $f$ be defined for all real $x$ , with $lim_{x \to 3} f (x) = 5$ and $f (3) = 7$ . Which statement must be true?
- A) $f$ is continuous at $x = 3$ , and $f$ is differentiable at $x = 3$
- B) $f$ is continuous at $x = 3$ , but $f$ is not differentiable at $x = 3$
- C) $f$ is not continuous at $x = 3$ , but $f$ is differentiable at $x = 3$
- D) $f$ is not continuous at $x = 3$ , and $f$ is not differentiable at $x = 3$
Reveal answer
D —
Continuity requires $lim_{x \to a} f (x) = f (a)$ . Since $5 \neq = 7$ , $f$ is discontinuous at $x = 3$ . By the contrapositive of differentiability implies continuity, a discontinuous function cannot be differentiable, so $f$ is also not differentiable at $x = 3$ .

5. Common Pitfalls

Wrong move:

After confirming derivatives of the two pieces of a piecewise function are equal at $x = a$ , conclude differentiability without checking continuity first.

Why:

Students assume matching derivatives imply the function is connected, skipping the required continuity check that is a necessary condition for differentiability.

Correct move:

Always check continuity at $x = a$ before comparing one-sided derivatives. If discontinuous, stop and conclude non-differentiability.

Wrong move:

Conclude a function is not continuous at $x = a$ just because it is not differentiable at $x = a$ .

Why:

Students confuse the direction of the implication, incorrectly inverting the 'differentiability implies continuity' rule.

Correct move:

Remember only 'discontinuous $\to$ not differentiable' is always true; non-differentiability does not imply discontinuity, so always test continuity separately.

Wrong move:

Count a removable discontinuity (hole) as differentiable because the limit of the function exists at the point.

Why:

Students confuse the limit of the function with the function being continuous at the point.

Correct move:

Any discontinuity (removable, jump, infinite) means the function is not continuous, hence not differentiable, regardless of whether the limit exists.

Wrong move:

Conclude $f (x) = ∣ x ∣$ is not continuous at $x = 0$ because it is not differentiable at $x = 0$ .

Why:

Students forget that $∣ x ∣$ is the classic example of a continuous non-differentiable function, and mix up the implication direction.

Correct move:

Check continuity first, then check differentiability: $∣ x ∣$ is continuous at $0$ , but non-differentiable.

Wrong move:

When comparing one-sided derivatives for a piecewise junction at $x = a$ , evaluate the derivatives at $x = 0$ instead of $x = a$ .

Why:

Students rush and default to plugging in the most common junction point, leading to an incorrect comparison.

Correct move:

Explicitly label the junction $x = a$ , write the derivative of each piece, and evaluate at $x = a$ before comparing.

6. Quick Reference Cheatsheet

Category	Rule/Formula	Notes
Differentiability Implies Continuity	If $f^{'} (a)$ exists, $f$ is continuous at $x = a$	Always true; contrapositive: discontinuous at $a$ → $f^{'} (a)$ does not exist
Converse Is False	Continuous at $a$ does not guarantee $f^{'} (a)$ exists	Continuity is necessary, not sufficient, for differentiability
Left-Hand Derivative	$f_{-}^{'} (a) = lim_{h \to 0^{-}} \frac{f ( a + h ) - f ( a )}{h}$	Equals derivative of left piece at $x = a$ for smooth pieces
Right-Hand Derivative	$f_{+}^{'} (a) = lim_{h \to 0^{+}} \frac{f ( a + h ) - f ( a )}{h}$	Equals derivative of right piece at $x = a$ for smooth pieces
Differentiability at Junction	$f$ differentiable at $a$ iff (1) continuous at $a$ , (2) $f_{-}^{'} (a) = f_{+}^{'} (a)$ finite	Check condition 1 first, always
Corner/Cusp	Non-differentiable	Continuous, but one-sided derivatives do not match
Vertical Tangent	Non-differentiable	Continuous, but slope is infinite (not finite)
Any Discontinuity	Non-differentiable	Applies to removable, jump, and infinite discontinuities

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · AP Calculus BC
MCQ check of implication direction
2022 · AP Calculus BC
FRQ piecewise differentiability check

What's Next

This topic is a foundational prerequisite for all subsequent differentiation work in AP Calculus BC. Next, you will learn the product rule, quotient rule, and chain rule for differentiating combinations of functions, all of which assume that the underlying functions are differentiable on their domains. You will also apply this relationship when you study derivatives of inverse functions, implicit differentiation, and the Mean Value Theorem, which explicitly requires a function to be continuous on a closed interval and differentiable on the open interval to apply. Without mastering the order of checking (continuity first, then differentiability) and the one-way nature of the implication, you will lose easy points on justification questions, which are graded strictly for correct logical reasoning on the AP exam.