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 always has the same shape: name the cause, then show the two one-sided slopes. "The graph has a sharp point" earns nothing without the numbers behind it.
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.
Figure 1 — Each panel shows one mechanism at the marked input. Blue is the backward quotient and amber is the forward one: at a corner they land on different finite numbers, at a cusp they run off in opposite directions, and at a vertical tangent they run off together.
| Cause | Example at x=0 | Continuous? | One-sided slopes | Failure mode |
|---|---|---|---|---|
| Discontinuity | f(x)=x1 | No | Not defined across the break | Continuity fails first |
| Corner | f(x)=∣x∣ | Yes | −1 and 1 | Two finite values differ |
| Cusp | f(x)=x2/3 | Yes | −∞ and +∞ | Divergence with opposite signs |
| Vertical tangent | f(x)=3x | Yes | +∞ and +∞ | 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.
At a=0 for f(x)=∣x∣, 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.
Both look dramatic, and the exam expects you to separate them. For x2/3 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 3x 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.
| 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 |
Borrowed idea: Read the shape. A derivative fails for four reasons, and each has a look. A corner has two slopes that disagree. A cusp has two slopes that run off to opposite infinities. A vertical tangent has two slopes that run off together. A break has no slope to measure, because the function value and the limit never meet. Match the shape to the reason before you write anything, and the justification writes itself in one sentence.
| 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)=∣x−1∣+∣x+1∣, 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.
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→0−, the difference quotient tends to −3.
- State the right slope. As h→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)=x2/3. Show that f is continuous at x=0 but not differentiable there, and identify the type of failure.
Solution: Continuity holds, since x2/3→0=f(0). For the quotients, hf(h)−f(0)=h∣h∣2/3, which behaves like h−1/3: as h→0− it tends to −∞ and as h→0+ it tends to +∞. Two infinite one-sided slopes with opposite signs make this a cusp, not a corner and not a vertical tangent.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
The graph of f has an unbroken curve with a sharp turn at x=1, where the left slope is 2 and the right slope is −2. Which statement is true?
MCQ 2
Let g(x)=3x. What happens to the difference quotient hg(h)−g(0) as h→0?
MCQ 3
The graph of f on [−3,4] has a jump at x=−2, a corner at x=0, and is smooth everywhere else. At how many inputs in (−3,4) does f′ fail to exist?
The graph of a continuous function f on [−4,4] has a corner at x=−1 with left slope 3 and right slope −1, a vertical tangent at x=1, and no other features.
(a) Explain why f′(−1) does not exist. Name the type of failure and support it with the two one-sided quotients.
(b) Explain why f′(1) does not exist, and state how the graph of f′ behaves near x=1.
(c) The function h is defined by h(x)=f(x)−2x. Is h differentiable at x=1? Justify your answer.
(d) A classmate claims that because f is continuous on [−4,4], f must be differentiable at every input except the two named above. Identify the flaw in that reasoning.
| Continuity | One-sided slopes | Graph of f′ there | |
|---|---|---|---|
| Discontinuity | Fails | Not defined | No graph, with an asymptote |
| Corner | Holds | Two finite values, unequal | A jump |
| Cusp | Holds | −∞ and +∞ | Asymptote with a sign flip |
| Vertical tangent | Holds | +∞ and +∞ | Asymptote without a sign flip |
Strengthen this topic with AP-style practice, or continue to next topic.
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 always has the same shape: name the cause, then show the two one-sided slopes. "The graph has a sharp point" earns nothing without the numbers behind it.
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.
Figure 1 — Each panel shows one mechanism at the marked input. Blue is the backward quotient and amber is the forward one: at a corner they land on different finite numbers, at a cusp they run off in opposite directions, and at a vertical tangent they run off together.
| Cause | Example at x=0 | Continuous? | One-sided slopes | Failure mode |
|---|---|---|---|---|
| Discontinuity | f(x)=x1 | No | Not defined across the break | Continuity fails first |
| Corner | f(x)=∣x∣ | Yes | −1 and 1 | Two finite values differ |
| Cusp | f(x)=x2/3 | Yes | −∞ and +∞ | Divergence with opposite signs |
| Vertical tangent | f(x)=3x | Yes | +∞ and +∞ | 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.
At a=0 for f(x)=∣x∣, 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.
Both look dramatic, and the exam expects you to separate them. For x2/3 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 3x 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.
| 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 |
Borrowed idea: Read the shape. A derivative fails for four reasons, and each has a look. A corner has two slopes that disagree. A cusp has two slopes that run off to opposite infinities. A vertical tangent has two slopes that run off together. A break has no slope to measure, because the function value and the limit never meet. Match the shape to the reason before you write anything, and the justification writes itself in one sentence.
| 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)=∣x−1∣+∣x+1∣, 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.
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→0−, the difference quotient tends to −3.
- State the right slope. As h→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)=x2/3. Show that f is continuous at x=0 but not differentiable there, and identify the type of failure.
Solution: Continuity holds, since x2/3→0=f(0). For the quotients, hf(h)−f(0)=h∣h∣2/3, which behaves like h−1/3: as h→0− it tends to −∞ and as h→0+ it tends to +∞. Two infinite one-sided slopes with opposite signs make this a cusp, not a corner and not a vertical tangent.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
The graph of f has an unbroken curve with a sharp turn at x=1, where the left slope is 2 and the right slope is −2. Which statement is true?
MCQ 2
Let g(x)=3x. What happens to the difference quotient hg(h)−g(0) as h→0?
MCQ 3
The graph of f on [−3,4] has a jump at x=−2, a corner at x=0, and is smooth everywhere else. At how many inputs in (−3,4) does f′ fail to exist?
The graph of a continuous function f on [−4,4] has a corner at x=−1 with left slope 3 and right slope −1, a vertical tangent at x=1, and no other features.
(a) Explain why f′(−1) does not exist. Name the type of failure and support it with the two one-sided quotients.
(b) Explain why f′(1) does not exist, and state how the graph of f′ behaves near x=1.
(c) The function h is defined by h(x)=f(x)−2x. Is h differentiable at x=1? Justify your answer.
(d) A classmate claims that because f is continuous on [−4,4], f must be differentiable at every input except the two named above. Identify the flaw in that reasoning.
| Continuity | One-sided slopes | Graph of f′ there | |
|---|---|---|---|
| Discontinuity | Fails | Not defined | No graph, with an asymptote |
| Corner | Holds | Two finite values, unequal | A jump |
| Cusp | Holds | −∞ and +∞ | Asymptote with a sign flip |
| Vertical tangent | Holds | +∞ and +∞ | Asymptote without a sign flip |
Strengthen this topic with AP-style practice, or continue to next topic.