Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist - AP Calculus AB

SECTION 1

Historical Context & Motivation

For most of the seventeenth and eighteenth centuries, mathematicians such as Newton and Leibniz operated under the informal assumption that every "reasonable" curve could be assigned a tangent line at every point. The notion of a derivative was synonymous with geometric smoothness, and no one questioned whether a continuous curve might fail to possess a tangent. It was not until the nineteenth century that rigorous definitions of continuity and differentiability exposed a surprising gap: a function can be perfectly continuous yet hopelessly non-differentiable. This realization reshaped analysis and eventually became one of the central distinctions tested on the AP Calculus AB exam.

1687 • Newton's Principia

Isaac Newton publishes the Principia, treating derivatives ("fluxions") as rates of change of smoothly varying quantities, with no formal distinction between continuity and differentiability.

1821 • Cauchy's Cours d'Analyse

Augustin-Louis Cauchy provides ε-δ style definitions for limits and continuity, laying the groundwork for rigorous differentiation theory.

1861 • Weierstrass's Monster

Karl Weierstrass presents a function that is continuous everywhere but differentiable nowhere—shattering the belief that continuity implies differentiability.

1872 • Formal Publication

Weierstrass formally publishes his continuous-but-nowhere-differentiable function, forcing the mathematical community to rigorously separate the two concepts.

The fundamental question this lesson addresses is: Under what conditions does a function possess a derivative, and what geometric or algebraic features cause the derivative to fail to exist? Mastering this distinction is critical not only for the AP exam but also for understanding the theoretical foundations upon which all of differential calculus rests.

SECTION 2

Core Principles & Definitions

Before exploring when derivatives do and do not exist, we must ground ourselves in precise definitions. Continuity at a point x = a requires three conditions: f(a) is defined, the limit of f(x) as x approaches a exists, and that limit equals f(a). Differentiability at x = a demands something stronger: the limit of the difference quotient [f(a + h) − f(a)] / h as h → 0 must exist as a single, finite real number. This limit, when it exists, defines f′(a).

Differentiability Implies Continuity

If f is differentiable at x = a, then f is necessarily continuous at x = a. This is a theorem, not an assumption—the proof follows directly from the limit definition of the derivative.

Continuity Does NOT Imply Differentiability

A function may be continuous at a point yet fail to be differentiable there. Corners, cusps, and vertical tangent lines are all continuous features where the derivative does not exist.

Left and Right Derivatives Must Agree

For f′(a) to exist, the left-hand derivative (limit as h → 0⁻) and the right-hand derivative (limit as h → 0⁺) must both exist and be equal. A mismatch signals a corner or cusp.

Four Classic Failure Modes

The derivative fails to exist at: discontinuities (removable, jump, or infinite), corners (sharp direction change), cusps (slopes diverge to ±∞), and vertical tangent lines (slope is ±∞).

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KEY TAKEAWAY

Think of differentiability as a stricter membership requirement than continuity. Being continuous is like having a ticket to enter a stadium—necessary but not sufficient. Differentiability is like having a ticket and a reserved seat assignment. Every person with a reserved seat (differentiable) is inside the stadium (continuous), but plenty of people inside the stadium are standing in the aisles without a designated seat (continuous but not differentiable).

SECTION 3

Visual Explanation: Where Derivatives Fail

The diagram below presents four scenarios where a function is continuous at a point but the derivative fails to exist. Each panel shows the graph of a function near a critical point, with the behavior of the tangent line (or lack thereof) illustrated. Study how the secant lines from the left and right approach different slopes in each case.

Four failure modes of differentiability. From left to right: a corner where left and right slopes disagree, a cusp where slopes diverge to ±∞, a vertical tangent where the slope is infinite, and a discontinuity which precludes differentiability entirely.

Pay special attention to the difference between the cusp and the vertical tangent. At a cusp, the one-sided derivatives diverge to opposite infinities (one to +∞ and the other to −∞), whereas at a vertical tangent both one-sided derivatives diverge to the same infinity (+∞ or −∞). In both cases the limit of the difference quotient is not a finite number, so f′(a) does not exist—but the geometric pictures are quite different, and AP Calculus AB questions may ask you to distinguish them.

SECTION 4

Mathematical Framework

The mathematical relationship between differentiability and continuity is anchored in the limit definition of the derivative. Every result in this section flows from that single definition, so we begin there and derive the fundamental theorem that differentiability implies continuity.

DEFINITION OF THE DERIVATIVE

f′(a) = lim(h→0) [f(a + h) − f(a)] / h

where a is the point of differentiation and h is the increment. The limit must exist as a single finite real number for f to be differentiable at a.

THEOREM: DIFFERENTIABILITY ⟹ CONTINUITY

If f′(a) exists, then lim(x→a) f(x) = f(a)

Proof sketch: Write f(x) − f(a) = [(f(x) − f(a))/(x − a)] · (x − a). As x → a, the first factor approaches f′(a) (a finite number) and the second factor approaches 0. Therefore lim(x→a)[f(x) − f(a)] = f′(a) · 0 = 0, which means lim(x→a) f(x) = f(a).

ONE-SIDED DERIVATIVE CONDITION

f′(a) exists ⟺ lim(h→0⁻) [f(a+h)−f(a)]/h = lim(h→0⁺) [f(a+h)−f(a)]/h ∈ ℝ

Both one-sided limits must exist, be finite, and be equal. If the left-hand and right-hand derivatives disagree (as at a corner), or if either diverges (as at a cusp or vertical tangent), the derivative does not exist.

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CONTRAPOSITIVE — THE EXAM SHORTCUT

Since differentiability implies continuity, the contrapositive is equally valid: if a function is not continuous at a, then it is automatically not differentiable at a. On the AP exam, always check continuity first. If the function is discontinuous, you can immediately conclude the derivative does not exist—no further computation needed.

SECTION 5

Detailed Classification of Non-Differentiable Points

The AP Calculus AB exam expects you to identify and describe points of non-differentiability by type. The following table and diagram provide a systematic classification. For each type, consider the behavior of the difference quotient from the left and from the right to understand exactly why the derivative fails to exist.

Type	Continuous?	Left Derivative	Right Derivative	Example
Corner	Yes	Finite value L₁	Finite value L₂ ≠ L₁	f(x) = \|x\| at x = 0
Cusp	Yes	+∞ (or −∞)	−∞ (or +∞)	f(x) = x²ᐟ³ at x = 0
Vertical Tangent	Yes	+∞ (or −∞)	+∞ (or −∞) same sign	f(x) = x¹ᐟ³ at x = 0
Jump Discontinuity	No	May exist finitely	May exist finitely	Piecewise step function
Removable Discontinuity	No	DNE (limit issue)	DNE (limit issue)	f(x) = (x²−1)/(x−1), x≠1

Classification of points where the derivative does not exist

A decision flowchart for determining differentiability. Always begin by checking continuity first, then verify that both one-sided derivatives exist and agree.

When approaching any AP question about differentiability, use this flowchart as a mental checklist. The first gate—continuity—is often the fastest to check and immediately eliminates a large class of problems. If the function passes the continuity test, compute the left-hand and right-hand derivatives separately using the limit definition. Only when both are finite and equal can you conclude that f′(a) exists.

SECTION 6

Worked Example

Let us work through a complete example that tests both continuity and differentiability for a piecewise function—a type frequently encountered on the AP Calculus AB exam.

Is the piecewise function differentiable at x = 2?

1
Step 1 — State the Function

Consider f(x) = { x² − 1 for x ≤ 2, and 3x − 3 for x > 2 }. We need to determine whether f is differentiable at x = 2.
2
Step 2 — Check Continuity at x = 2

Compute f(2) from the first piece: f(2) = 2² − 1 = 3. Compute the left-hand limit: lim(x→2⁻) (x² − 1) = 4 − 1 = 3. Compute the right-hand limit: lim(x→2⁺) (3x − 3) = 6 − 3 = 3. Since f(2) = lim(x→2⁻) f(x) = lim(x→2⁺) f(x) = 3, the function is continuous at x = 2.

Continuous at x = 2 ✓
3
Step 3 — Compute the Left-Hand Derivative

Using the piece f(x) = x² − 1 for x ≤ 2: f′₋(2) = lim(h→0⁻) [f(2+h) − f(2)] / h = lim(h→0⁻) [(2+h)² − 1 − 3] / h = lim(h→0⁻) [4 + 4h + h² − 4] / h = lim(h→0⁻) [4h + h²] / h = lim(h→0⁻) (4 + h) = 4.

Left-hand derivative = 4
4
Step 4 — Compute the Right-Hand Derivative

Using the piece f(x) = 3x − 3 for x > 2: f′₊(2) = lim(h→0⁺) [f(2+h) − f(2)] / h = lim(h→0⁺) [3(2+h) − 3 − 3] / h = lim(h→0⁺) [6 + 3h − 6] / h = lim(h→0⁺) 3h / h = 3.

Right-hand derivative = 3
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Step 5 — Compare and Conclude

The left-hand derivative (4) does not equal the right-hand derivative (3). Therefore the two-sided limit of the difference quotient does not exist. The function f is continuous at x = 2 but not differentiable there—this is a corner.

f′(2) does not exist (corner)

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AP EXAM TIP

On free-response questions, you earn separate points for verifying continuity, computing each one-sided derivative, and stating a clear conclusion with justification. Even if the final answer is incorrect, showing all three steps earns partial credit.

SECTION 7

Common Pitfalls & Comparisons

Students frequently confuse related but distinct concepts when determining differentiability. The table below contrasts common misconceptions with the correct reasoning, and highlights the most frequent errors that appear on the AP exam.

Common Misconception	Correct Understanding
"If f is continuous at a, then f is differentiable at a."	Continuity is necessary but not sufficient. f(x) = \|x\| is continuous everywhere but not differentiable at x = 0.
"Piecewise functions are never differentiable at breakpoints."	If both pieces give the same function value and the same derivative at the breakpoint, the function is differentiable there. Example: f(x) = {x² for x ≤ 1, 2x − 1 for x > 1} is differentiable at x = 1.
"A vertical tangent means the function is discontinuous."	A vertical tangent (like at x = 0 for f(x) = x¹ᐟ³) is a feature of a continuous function. The derivative fails to exist because the slope is infinite, not because of a discontinuity.
"If f′(a) doesn't exist, there's no tangent line at a."	At a vertical tangent there is a tangent line—it is simply vertical (undefined slope). At corners, there are two distinct tangent directions, not zero.
"Differentiability only matters at integer or "nice" values."	Differentiability can fail at any x-value. Always examine points where the function formula changes or where algebraic expressions involve absolute values, fractional exponents, or radicals.

Common misconceptions versus correct reasoning

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KEY TAKEAWAY

Think of the relationship like this: being a licensed driver (differentiable) requires you to be at least 16 years old (continuous). However, many people who are 16 or older (continuous) are not licensed drivers (differentiable)—perhaps they never took the test, or their license was revoked. The age requirement is necessary but not sufficient, just as continuity is necessary but not sufficient for differentiability.

SECTION 8

Connections to Advanced Theory

The differentiability–continuity relationship you have studied in this lesson extends into more sophisticated mathematical territory that you will encounter in AP Calculus BC and beyond. Understanding these connections now will deepen your intuition and prepare you for future coursework.

AP Calculus AB Concept	Advanced Extension	Key Idea
Differentiability implies continuity	Continuously differentiable functions (C¹)	In analysis, we classify functions by how many times they are continuously differentiable. C¹ means f′ exists and is itself continuous.
Corners and cusps (non-differentiable points)	Lipschitz continuity and weak derivatives	\|x\| has a weak derivative in the sense of distributions, even though the classical derivative fails at 0. This idea powers PDEs and real analysis.
Weierstrass continuous-but-nowhere-differentiable function	Fractal geometry and Brownian motion	Many natural phenomena (stock prices, coastlines) follow paths that are continuous but nowhere differentiable—an idea central to fractal mathematics.
One-sided derivatives	Directional derivatives in multivariable calculus	The left/right derivative generalizes to directional derivatives in ℝⁿ, where differentiability requires the existence of a linear approximation in all directions simultaneously.

From AB fundamentals to advanced theory

For the AP Calculus AB exam, you do not need to master these advanced topics, but being aware of them reinforces a crucial lesson: the seemingly simple theorem that differentiability implies continuity is the first rung on a very tall ladder. Each successive course in analysis adds conditions—continuous first derivative, continuous second derivative, and so on—that control the smoothness of functions with increasing precision. The one-sentence version of the AB-level takeaway remains: differentiability is a stronger condition than continuity, and you must verify it by checking one-sided limits of the difference quotient.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Which of the following statements is true about the relationship between differentiability and continuity?

A. If f is continuous at x = a, then f is differentiable at x = a.
B. If f is differentiable at x = a, then f is continuous at x = a.
C. If f is not differentiable at x = a, then f is not continuous at x = a.
D. Differentiability and continuity are equivalent conditions.
E. A function can be differentiable at a point where it is not continuous.

Show answer (B)

The correct answer is (B). The theorem "differentiability implies continuity" guarantees that if f′(a) exists, then f must be continuous at a. Choice (A) is the converse, which is false—f(x) = |x| is continuous at x = 0 but not differentiable there. Choice (C) is the inverse, also false for the same reason: a function can be non-differentiable yet continuous (e.g., at a corner). Choice (D) claims equivalence, which contradicts all known counterexamples. Choice (E) directly contradicts the theorem, since differentiability requires continuity as a prerequisite.

Question 2: BASIC CALCULATION

Let f(x) = |x − 3|. At which point is f continuous but not differentiable?

A. x = 0
B. x = 1
C. x = 3
D. x = −3
E. f is differentiable everywhere

Show answer (C)

The correct answer is (C). The function f(x) = |x − 3| can be rewritten as a piecewise function: f(x) = −(x − 3) for x < 3 and f(x) = x − 3 for x ≥ 3. The function is continuous everywhere, including at x = 3. However, at x = 3, the left-hand derivative is −1 and the right-hand derivative is +1. Since these are unequal, f is not differentiable at x = 3 (a corner). Choices (A), (B), and (D) are incorrect because at those points f is a simple linear function and hence differentiable. Choice (E) is incorrect because the corner at x = 3 prevents differentiability there.

Question 3: INTERMEDIATE

Consider the piecewise function f(x) = { x² + 1 for x ≤ 1, and 2x + k for x > 1 }. Find the value of k that makes f differentiable at x = 1. If no such value exists, explain why.

A. k = 0
B. k = 1
C. k = −1
D. k = 2
E. No value of k makes f differentiable at x = 1

Show answer (A)

The correct answer is (A). For differentiability we need both continuity and matching one-sided derivatives. First, continuity: f(1) = 1² + 1 = 2 from the left piece. The right-hand limit is lim(x→1⁺)(2x + k) = 2 + k. Setting 2 + k = 2 gives k = 0. Now check derivatives: the left-hand derivative is f′(x) = 2x evaluated at x = 1, giving 2. The right-hand derivative is f′(x) = 2 for all x > 1, giving 2. Since both one-sided derivatives equal 2 and continuity holds with k = 0, the function is differentiable at x = 1. Choice (B) would give a right-hand limit of 3, breaking continuity. Choice (C) gives a right-hand limit of 1, also breaking continuity. Choice (D) gives a right-hand limit of 4, again not continuous. Choice (E) is incorrect because k = 0 satisfies all requirements.

Question 4: APPLIED

5 pts

A particle moves along the x-axis with position function s(t) = { t² for 0 ≤ t < 2, and 4t − 4 for t ≥ 2 }. (a) Is s(t) continuous at t = 2? Justify your answer. (b) Is s(t) differentiable at t = 2? Justify your answer by computing the left-hand and right-hand derivatives. (c) What is the physical significance of your answer to part (b) in terms of the particle's velocity?

Show answer

Point 1 (Continuity check): s(2) = 2² = 4 from the left piece. lim(t→2⁻) t² = 4 and lim(t→2⁺)(4t − 4) = 8 − 4 = 4. Since all three values equal 4, s is continuous at t = 2. Point 2 (Left-hand derivative): s′₋(2) = lim(h→0⁻)[(2+h)² − 4]/h = lim(h→0⁻)[4 + 4h + h² − 4]/h = lim(h→0⁻)(4 + h) = 4. Point 3 (Right-hand derivative): s′₊(2) = lim(h→0⁺)[(4(2+h) − 4) − 4]/h = lim(h→0⁺)[8 + 4h − 4 − 4]/h = lim(h→0⁺) 4h/h = 4. Point 4 (Differentiability conclusion): Since the left-hand derivative (4) equals the right-hand derivative (4), and the function is continuous at t = 2, the function is differentiable at t = 2 with s′(2) = 4. Point 5 (Physical interpretation): Because s′(2) = 4 exists, the particle's instantaneous velocity at t = 2 is 4 units per second. There is no abrupt change in velocity at t = 2; the particle transitions smoothly between the two motion regimes. If the derivatives had not matched, the particle would experience an instantaneous jump in velocity, which would be physically unrealizable for a particle with finite mass.

Question 5: CRITICAL THINKING

4 pts

Let g(x) = x · sin(1/x) for x ≠ 0 and g(0) = 0. (a) Prove that g is continuous at x = 0. (b) Determine whether g is differentiable at x = 0 by evaluating the limit of the difference quotient. (c) Explain what this example reveals about the relationship between continuity and differentiability.

Show answer

Point 1 (Continuity proof): Since −1 ≤ sin(1/x) ≤ 1, we have −|x| ≤ x · sin(1/x) ≤ |x| for all x ≠ 0. By the Squeeze Theorem, lim(x→0) x · sin(1/x) = 0 = g(0), so g is continuous at x = 0. Point 2 (Difference quotient setup): g′(0) = lim(h→0) [g(h) − g(0)]/h = lim(h→0) [h · sin(1/h)]/h = lim(h→0) sin(1/h). Point 3 (Evaluating the limit): As h → 0, sin(1/h) oscillates between −1 and 1 without approaching any single value. Therefore lim(h→0) sin(1/h) does not exist, and g′(0) does not exist. Point 4 (Conceptual explanation): This example demonstrates that continuity does not imply differentiability in a subtle way—there is no corner, cusp, or vertical tangent. Instead, the function oscillates so rapidly near x = 0 that the secant line slopes never settle on a single value. It shows that the failure of differentiability can arise from oscillatory behavior, not just from the geometric features typically emphasized in AP Calculus. This reinforces that differentiability is a strictly stronger condition than continuity.

SUMMARY

Lesson Summary

Summary

This lesson established the foundational relationship between differentiability and continuity: if a function is differentiable at a point, it must be continuous there, but the converse is false. We proved this theorem using the limit definition of the derivative and its powerful contrapositive: if a function is not continuous, it cannot be differentiable.

We classified four types of non-differentiable points: corners (finite but unequal one-sided derivatives), cusps (one-sided derivatives diverge to opposite infinities), vertical tangent lines (one-sided derivatives diverge to the same infinity), and discontinuities (continuity fails, so differentiability is precluded). For the AP exam, always follow the decision flowchart: check continuity first, then compute and compare the left-hand and right-hand derivatives.