Unit 2Non-Differentiable Points

Concept 4.115Built for May 2027 AP Exam

Non-Differentiable Points

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

A point of non-differentiability is a place where the curve has no single tangent slope. Four causes cover every case the exam uses. A discontinuity of any kind ends the analysis immediately, because continuity is a precondition. A corner gives two finite one-sided slopes that disagree. A cusp gives two one-sided slopes that run to opposite infinities. A vertical tangent gives two one-sided slopes that run to the same infinity. The exam asks you to count these points on a graph or to justify one at a piecewise join, and the justification is always the same shape: name the cause and show the two one-sided slopes. "The graph has a sharp point" earns nothing without the numbers behind it.

On the AP Exam

Scoring here is about the reason, not the verdict. Saying "not differentiable" is worth nothing on its own; the point lives in the sentence that follows it. The grader is checking whether you identified the mechanism, and the four mechanisms are not interchangeable, because a corner needs two finite slopes while a cusp needs two infinite ones.

What you’ll see: A graph of $f$ with marked points, a piecewise formula with a join to test, or a question asking how many points on an interval fail to be differentiable.
What you’ll do: Test continuity first, then compare the one-sided difference quotients at every remaining candidate.
Memory hook: Break beats bend. A break kills differentiability outright; a bend is the case you have to measure.
Use it on the exam: For MCQs: Count the candidates left to right, and mark each one with the cause before deciding. Corners on a graph of $f$ become jumps on the graph of $f'$ , which is often the fastest way to eliminate an option. For FRQs: One released part, 2017 Q3(d), asks why $f''(3)$ fails to exist, and the accepted reason is that the two one-sided difference quotients settle at different finite values. Quote the two numbers, then conclude.

Concept Review

Figure 1

Each card shows what the one-sided difference quotients do at the input in question. A break removes the comparison entirely, which is why continuity is tested before any slope is computed.

Four cards listing the causes of non-differentiability with the one-sided slope behaviour each produces.

The Four Causes

Cause	Example at $x = 0$	Continuity	One-sided slopes	Failure mode
Discontinuity	$f(x) = \frac{1}{x}$	No	Not defined across the break	Continuity fails first
Corner	$f(x) = \lvert x \rvert$	Yes	$-1$ and $1$	Two finite values differ
Cusp	$f(x) = x^{2/3}$	Yes	$-\infty$ and $+\infty$	Divergence with opposite signs
Vertical tangent	$f(x) = \sqrt[3]{x}$	Yes	$+\infty$ and $+\infty$	Divergence in the same direction

The table is ordered by cost. If continuity fails, nothing else needs checking. If continuity holds, the three remaining causes are distinguished by what the one-sided quotients do: land on different numbers, diverge apart, or diverge together.

Why a Corner Fails

At $a = 0$ for $f(x) = \lvert x \rvert$ , the difference quotient is $-1$ for every negative $h$ and $1$ for every positive $h$ . The two one-sided limits exist and disagree, so the two-sided limit does not exist, and the derivative is undefined. Nothing is infinite and nothing is missing; the slope simply jumps, which is why the graph of $f'$ shows a jump at that input.

Why a Cusp and a Vertical Tangent Differ

Both look dramatic, and the exam expects you to separate them. For $x^{2/3}$ the quotient behaves like $h^{-1/3}$ , which is negative for $h < 0$ and positive for $h > 0$ , so the two sides head to opposite infinities. For $\sqrt[3]{x}$ the quotient behaves like $h^{-2/3}$ , which is positive on both sides, so the two infinities agree in sign. A vertical tangent is a single infinite slope; a cusp is a disagreement between two infinite slopes.

Reading Failures off the Graph of the Derivative

On the graph of $f$	On the graph of $f'$	Reading
Jump in $f$	No graph at that input	$f'$ undefined, and the gap is a vertical asymptote
Corner	Jump from one finite value to another	$f'$ has a discontinuity with finite one-sided limits
Cusp	Behaviour running to opposite infinities	$f'$ has a vertical asymptote with a sign flip
Vertical tangent	Behaviour running to the same infinity	$f'$ has a vertical asymptote without a sign flip

Common Misconception Check

Myth 1: "Only sharp points cause non-differentiability." Truth: Every discontinuity is also a failure, and a smooth-looking curve with a vertical tangent fails as well.
Myth 2: "A cusp and a corner are the same thing." Truth: At a corner both one-sided slopes are finite and unequal; at a cusp both are infinite, with opposite signs.
Myth 3: "If the graph looks smooth, the derivative exists." Truth: Smoothness is a drawing convention. The test is a limit of quotients, and $x^{1/3}$ passes the eye test while failing the limit test.

Topic-Specific Study Method

Name It Before You Prove It

Borrowed idea: Non-differentiability is a diagnosis with four causes, and the picture usually tells you which one you are looking at. A break in the curve settles the question before any limit gets written down. An unbroken curve with a sharp turn gives two finite one-sided slopes that disagree. A pointed tip where the curve turns back sends the two one-sided slopes to opposite infinities. A vertical stretch sends both of them to the same infinity. Name the cause first, because the name dictates the sentence the grader is looking for.

Exam disguise	What you see on the page	What your brain must translate	Required execution
"How many points fail?"	A curve with several marked features	One question, four possible causes	Scan for breaks first, then bends
A piecewise formula	Two expressions joined at $x = c$	Continuity and slope are separate tests	Compare values, then compare quotients
A graph of $f'$	A jump in the derivative	A corner in $f$ was hidden in the picture	Report the input where the jump sits
"Explain why $f'(0)$ does not exist"	A single point named	The reason must be quantitative	Compute both one-sided quotients

① On $f(x) = \lvert x - 1 \rvert + \lvert x + 1 \rvert$ , find every input where the derivative fails and classify the cause. Two corners appear, and both are finite-slope failures.

② Rank the four causes by how quickly you can detect them from a picture, then check that ranking against the algebra. Breaks are instant, corners need one comparison, cusps and vertical tangents need a limit.

③ When a rubric asks for a justification, write the two one-sided quotients as numbers or as infinities before writing the conclusion. A conclusion without numbers is a claim; a conclusion after numbers is a proof.

Training & Error Diagnosis

Core Archetype (Simplest Form): The graph of $f$ has a corner at $x = 2$ , where the left slope is $-3$ and the right slope is $4$ . Explain why $f'(2)$ does not exist.

Standard Execution:

Confirm continuity. The two branches meet, so the function value is shared and continuity holds.

State the left slope. As $h \to 0^-$ , the difference quotient tends to $-3$ .

State the right slope. As $h \to 0^+$ , the difference quotient tends to $4$ .

Compare. Two finite limits that differ, so the two-sided limit does not exist.

Conclude. Therefore $f'(2)$ is undefined, and the graph of $f'$ jumps from $-3$ to $4$ at that input.

Exam Variation (Context / Parameter Shift): Let $f(x) = \begin{cases} x^2 & x < 1 \cr 2x - 1 & x \ge 1 \end{cases}$ . Determine whether $f$ is differentiable at $x = 1$ .

Solution: Continuity first: the left limit is $1$ and the right value is $2(1) - 1 = 1$ , so $f$ is continuous at $1$ . Now the quotients. On the left, $\frac{x^2 - 1}{x - 1} = x + 1 \to 2$ . On the right, the branch is a line of slope $2$ . The two one-sided slopes both equal $2$ , so $f'(1) = 2$ and $f$ is differentiable at $1$ despite the change of formula. A formula change is not a corner; only a slope change is.

Common Traps & Failure Signals:

The verdict without the reason (Signal: "not differentiable" with no quotients shown): the conclusion was written, the evidence was not.
The premature slope test (Signal: quotients compared on a function that was never checked for continuity): the order was reversed, so the justification rests on an untested assumption.
The mislabelled tip (Signal: the word cusp used for any sharp point): the category was chosen by appearance rather than by the sign of the infinite limits.

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

Discontinuity Blind Spot $\rightarrow$ Check continuity before you hunt for corners.
One-Sided Skip: You compared the heights but never the slopes.
Category Confusion: You wrote "cusp" for a corner and lost the justification point, because the two failures look alike in a small picture but require opposite one-sided limits, and the scoring guideline checks which name you attached to the evidence.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; One scored part asks why a derivative fails to exist at a marked input, and the accepted reason is a disagreement between two one-sided difference quotients
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part
FRQ Years Appeared: 2017 (Q3d)

Practice Examples

MCQ 1

Counting the Failures

The graph of $f$ consists of a line segment from $(-3, 1)$ to $(0, 1)$ , a line segment from $(0, 1)$ to $(2, 3)$ , and a semicircle from $(2, 3)$ to $(4, 3)$ that opens downward. At how many inputs in $(-3, 4)$ does $f'$ fail to exist?

MCQ 2

The Cusp Test

Let $f(x) = x^{2/3}$ . Which statement describes $f$ at $x = 0$ ?

MCQ 3

The Piecewise Join

Let $g(x) = \begin{cases} 3x + 2 & x \le 1 \cr x^2 + 4 & x > 1 \end{cases}$ . Which statement is true?

FRQ 4

Two Joins on One Graph

The graph of the continuous function $f$ on $[-3, 4]$ consists of a quarter circle of radius $2$ centered at $(-1, 0)$ from $x = -3$ to $x = -1$ , a line segment from $(-1, 2)$ to $(2, 2)$ , and a line segment from $(2, 2)$ to $(4, -2)$ .

(a) Find every input in $(-3, 4)$ at which $f'$ does not exist, and classify the cause at each one.

(b) At $x = -1$ , determine whether $f$ is differentiable. Justify your answer with one-sided slopes.

(c) Sketch a description of the graph of $f'$ on $(-3, 4)$ , naming the behaviour at each input found in part (a).

(d) Explain why the fact that $f$ is continuous on $[-3, 4]$ does not guarantee that $f'$ exists at every input in $(-3, 4)$ .

Cheatsheet