Unit 2Differentiability Implies Continuity

Concept 4.138Built for May 2027 AP Exam

Differentiability Implies Continuity

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Differentiability is the stronger property. If $f$ has a derivative at $a$ , then $f$ is continuous at $a$ , and the reason is structural rather than case-by-case: the difference quotient can only converge if its numerator shrinks at least as fast as its denominator. The implication runs one way only. A continuous function may fail to have a derivative at a point, and every corner, cusp, vertical tangent, or break is a witness. On the exam this appears as a statement to judge rather than a computation to finish. You are asked whether differentiability follows from continuity, or whether a theorem that needs continuity may be applied to a function described only as differentiable. The contrapositive is the working tool: a visible break settles non-differentiability at once, with no limit to compute.

On the AP Exam

This item is scored on logic, not on algebra. The grader looks for one thing: whether the direction of the arrow in your sentence matches the direction the hypothesis allows. Because nothing needs computing, points are lost inside a single word, usually by writing "continuous, therefore differentiable" in a justification that was otherwise correct.

What you’ll see: A statement to classify as always, sometimes, or never true; a piecewise function with a join to test; or a theorem that needs continuity as a precondition.
What you’ll do: Name which of the two properties you were given, name which one you were asked to supply, and check that the arrow points the right way.
Memory hook: Differentiable is the expensive one. It buys continuity at the same point, and it buys nothing in the other direction.
Use it on the exam: For MCQs: Read the statement as an implication and test it against a single counterexample, usually $f(x) = |x|$ at the origin. One counterexample kills a universal claim. For FRQs: In nine released exams no scored part asks you to prove the implication itself. What the rule buys you is a sentence: a function given as differentiable may be called continuous on the closed interval, which is what the Mean Value Theorem requires.

Concept Review

Figure 1

The solid arrow is the theorem and the dashed one is the error: differentiability forces continuity, and no amount of continuity produces a derivative. The three chips at the bottom are continuous functions that fail at a single input.

A solid arrow runs from differentiable to continuous; a crossed-out dashed arrow runs back. Three chips list corner, cusp, and vertical tangent as continuous functions with no derivative.

Four Statements, Two Truths

Statement	Symbolic form	Verdict
Differentiable at $a$ implies continuous at $a$	$D \Rightarrow C$	True
Continuous at $a$ implies differentiable at $a$	$C \Rightarrow D$	False
Not continuous at $a$ implies not differentiable at $a$	$\lnot C \Rightarrow \lnot D$	True — the contrapositive of row 1
Not differentiable at $a$ implies not continuous at $a$	$\lnot D \Rightarrow \lnot C$	False — the inverse, not the contrapositive

Rows 1 and 3 say the same thing. Rows 2 and 4 say the same wrong thing. An implication and its contrapositive are logically identical; an implication and its inverse are not, and the inverse is the error almost every student makes at least once.

Why the Implication Holds

Write the difference quotient and clear its denominator:

$\frac{f(a+h)-f(a)}{h} = Q(h) \qquad\text{so}\qquad f(a+h)-f(a) = h \cdot Q(h)$

If $f$ is differentiable at $a$ , then $Q(h) \to f'(a)$ as $h \to 0$ , while $h \to 0$ on its own. The product of something converging to a finite number and something converging to zero converges to zero, so

$\lim_{h \to 0}\bigl[f(a+h)-f(a)\bigr] = 0 \qquad\text{which is}\qquad \lim_{h\to 0} f(a+h) = f(a)$

That last line is continuity at $a$ . Nothing in the argument survives if $Q(h)$ fails to converge, which is exactly why the converse collapses: the algebra needs a limit that may not exist.

Three Witnesses Against the Converse

Function	Point	Continuous?	Differentiable?	What fails
$f(x) = \lvert x \rvert$	$x = 0$	Yes	No	One-sided quotients settle at $-1$ and $1$
$f(x) = x^{2/3}$	$x = 0$	Yes	No	One-sided quotients diverge with opposite signs
$f(x) = \sqrt[3]{x}$	$x = 0$	Yes	No	One-sided quotients both diverge to $+\infty$

Each function is unbroken at the point in question, and each fails the derivative for a different reason. That variety is the point: continuity controls whether the graph has a hole or a jump, and it says nothing about whether the graph has a corner.

Where the Continuity Is Actually Spent

The theorem is rarely the answer; it is the licence for the next step. Three places the licence gets used:

Mean Value Theorem. Requires continuity on $[a,b]$ and differentiability on $(a,b)$ . When a problem grants differentiability on the closed interval, the continuity hypothesis is already met, and saying so costs one clause.
Extreme Value Theorem. Requires continuity on a closed interval. A function described as differentiable on $[-3, 4]$ therefore attains both a maximum and a minimum there.
L'Hospital's Rule. Requires the numerator and denominator to be differentiable near the point, which silently supplies continuity in a neighbourhood.

Common Misconception Check

Myth 1: "The graph has no break, so the derivative exists." Truth: No break gives continuity. A corner is unbroken and still has no derivative, because the two one-sided quotients disagree.
Myth 2: "If $f$ is not differentiable at $a$ , then $f$ is not continuous at $a$ ." Truth: This is the inverse, not the contrapositive. Non-differentiability may come from a corner, which is continuous.
Myth 3: "Continuity at every point implies differentiability at every point." Truth: Weierstrass built continuous functions that are differentiable nowhere. On the exam, a single corner on a continuous curve is enough to defeat the claim.

Topic-Specific Study Method

The Direction Check

Borrowed idea: Which direction does the arrow point? Differentiability is the harder property, so it hands continuity over for free, while continuity hands nothing back. So what does continuity actually buy you? It buys the Intermediate Value Theorem and the Extreme Value Theorem, and it never buys a derivative. When a problem gives you differentiability, spend it at once: write the continuity sentence and collect the theorem you need. A function that is differentiable everywhere except at one input is still continuous there, and that single gap in the domain of $f'$ is exactly what a later part of the question will ask you to describe.

Exam disguise	What you see on the page	What your brain must translate	Required execution
A true/false statement	"If $f$ is continuous at $a$ , then $f'(a)$ exists"	The converse, stated as fact	Reject it with the corner counterexample
A piecewise join	Two formulas meeting at $x = 2$	Continuity and differentiability are two separate tests	Run both tests, in that order
A theorem setup	"The function is differentiable on $[0, 5]$ "	Continuity is a gift you should claim	Write the continuity clause before quoting the theorem
A graph with a sharp turn	An unbroken curve with a point	Break rules out differentiability, a corner does not	Check the one-sided slopes, not just the height

① Take $f(x) = \lvert x - 2 \rvert$ and answer two separate questions: is $f$ continuous at $2$ , and is $f$ differentiable at $2$ ? Write both answers in full sentences, and notice that the first is a limit of values while the second is a limit of quotients.

② For any piecewise definition in your practice set, compute the two one-sided limits of $f$ at the join, then compute the two one-sided limits of the difference quotient. Two pairs of numbers, two verdicts, and the order matters: a failed continuity test ends the analysis immediately.

③ Whenever a solution you are reading says "since $f$ is differentiable", insert in the margin what that sentence is buying: continuity, and which theorem needs it. If nothing downstream needs it, the sentence was decoration.

Training & Error Diagnosis

Core Archetype (Simplest Form): A function $f$ is differentiable at $x = 3$ . A classmate concludes that $f'(3)$ is positive because the graph passes through the point $(3, 5)$ . Identify the error.

Standard Execution:

Separate the two claims. Differentiability is a statement about a limit existing, not about its sign.

Name what is guaranteed. Continuity at $3$ , and the existence of a finite $f'(3)$ .

Name what is not. The sign of $f'(3)$ is unconstrained; the derivative may be negative, zero, or positive.

Name the second error. The height $f(3) = 5$ carries no information about the slope at $3$ .

Repair the sentence. "Differentiability at $3$ guarantees that $f$ is continuous at $3$ and that $f'(3)$ is a finite number."

Exam Variation (Context / Parameter Shift): Let $f(x) = \begin{cases} x^2 + 1 & x \le 2 \cr ax + b & x > 2 \end{cases}$ . Find the values of $a$ and $b$ that make $f$ differentiable at $x = 2$ .

Solution: Continuity comes first. The left value is $f(2) = 5$ and the right-hand limit is $2a + b$ , so continuity requires $2a + b = 5$ . Differentiability comes second. The left-hand slope is $2x$ evaluated at $2$ , which is $4$ ; the right-hand slope is $a$ . Equal one-sided slopes require $a = 4$ , and the continuity equation then gives $b = -3$ . The order is not optional: differentiability presupposes continuity, so a pair that satisfies only the slope condition would be rejected.

Common Traps & Failure Signals:

The converse swap (Signal: the word "therefore" joining continuity to differentiability): the implication was used backwards.
The single-test stop (Signal: slopes compared but values never checked): the join was tested for differentiability without testing continuity, and the answer passed a condition that was never verified.
The unclaimed hypothesis (Signal: a theorem quoted with no continuity sentence): differentiability was granted and never converted, leaving a gap the grader is instructed to look for.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Converse Shortcut $\rightarrow$ Name the direction before you use the statement.
Silent Assumption: You called the curve differentiable because it looked unbroken.
Unclaimed Hypothesis: You were handed differentiability and never converted it into the continuity that the Mean Value Theorem needs, so the justification for applying the theorem on a closed interval went missing at exactly the step the grader was reading.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; The implication is judged inside multiple-choice statements; no released free-response part asks for the argument or for the proof
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; All 54 free-response questions from the nine released sets were read part by part
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Reading the Implication

The function $f$ is differentiable at $x = 2$ . Which statement must be true?

MCQ 2

The Counterexample

Let $f(x) = \lvert x - 3 \rvert$ . Which statement is true at $x = 3$ ?

MCQ 3

The Contrapositive

The graph of $g$ has a jump discontinuity at $x = 1$ . Which conclusion is justified?

FRQ 4

A Piecewise Join

Let $f$ be the function defined by

$f(x) = \begin{cases} 2x + 3 & x \le 1 \cr x^2 + kx + 4 & x > 1 \end{cases}$

where $k$ is a constant.

(a) Find the value of $k$ for which $f$ is continuous at $x = 1$ . Show the work that leads to your answer.

(b) Using the value of $k$ from part (a), determine whether $f$ is differentiable at $x = 1$ . Justify your answer.

(c) Explain why a function that is not continuous at a point cannot be differentiable there.

(d) The function $g$ is differentiable on $[-2, 5]$ with $g(-2) = 3$ and $g(5) = 3$ . Explain why $g$ must attain a maximum value on $[-2, 5]$ , and state which theorem supplies the guarantee.

Cheatsheet