Unit 5MVT Statement

Concept 4.231Built for May 2027 AP Exam

MVT Statement

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

The Mean Value Theorem converts an average into an instant. If a function is continuous on a closed interval and differentiable inside it, then somewhere inside the interval the instantaneous rate equals the average rate over the whole interval. Geometrically, some tangent line is parallel to the secant line joining the endpoints. The theorem is an existence claim, so it never tells you where that point is, only that at least one exists. It also needs both hypotheses, and a single corner or a single vertical asymptote is enough to break the conclusion. On the exam the work is almost always the same three moves: name the theorem, verify the conditions, then quote the conclusion with an interval attached.

On the AP Exam

What you’ll see: A question that asks you to show some rate must have occurred, phrased as "is there a time $t$ " or "must there exist a value $c$ ", usually with a table of values instead of a formula.
What you’ll do: Compute the average rate over the stated interval, state that the function is continuous and differentiable, then invoke the theorem to produce a $c$ with $f'(c)$ equal to that average.
Memory hook: Average in, instant out. The average rate is the input, and the theorem hands back an instant where that rate happens.
Use it on the exam: For MCQs: The distractors usually offer a specific value of $c$ , which the theorem cannot supply, or they let the theorem apply on an interval where differentiability fails. For FRQs: On 2021 Q4(d) the guideline gives a point for the average rate and a separate point for the correct conclusion about whether the theorem guarantees such a $c$ , and a response that only computes the average loses half of it.

Concept Review

The Statement, Split into Hypotheses and Conclusion

Figure 1 — The secant through the endpoints and a parallel tangent somewhere inside. The theorem guarantees that the parallel tangent exists; it says nothing about how many such points there are or where they sit.

Let $f$ satisfy two conditions: continuous on $[a,b]$ , differentiable on $(a,b)$ . Then there is at least one $c$ in $(a,b)$ with

$f'(c) = \frac{f(b)-f(a)}{b-a}$

The right side is the average rate of change, which is the slope of the secant line. The left side is the slope of the tangent at $c$ . The theorem says those two numbers coincide somewhere strictly inside the interval, never at an endpoint.

A concave down curve with its secant line between two endpoints and a parallel tangent line touching the curve at an interior point c

What the Theorem Does and Does Not Promise

It promises existence, not location and not uniqueness. A cubic can have two points with the secant slope, and the theorem is satisfied by either of them. It also says nothing when the interval is a single point, so the words "open" and "closed" in the statement are doing real work. A response that writes "the theorem gives $c = 2$ " has invented information.

Reading It as an Average-to-Instant Translation

Average rate is what a table gives you: change in output divided by change in input. Instantaneous rate is what a derivative gives you. The theorem is the bridge, and every applied version of it is the same bridge:

$\text{average rate over }[a,b] \;=\; \text{some } f'(c)$

That is why the exam asks the question with a table. The table supplies the average, the derivative supplies the instant, and the theorem supplies the guarantee that the two meet.

The Two Ways the Hypotheses Fail

A corner. For $f(x)=\lvert x\rvert$ on $[-1,1]$ , the average rate is $0$ , but no tangent has slope $0$ , because the derivative does not exist at $0$ . Continuity alone is not enough.
A vertical asymptote. For $f(x)=1/x$ on $[-1,1]$ , the function is not even continuous on the closed interval, so the theorem is silent and the average rate is meaningless.

Both failures are drawn from the same idea. The hypotheses are exactly the conditions that rule out the two ways a curve can refuse to have a parallel tangent.

The Consequences You Will Use Later

The theorem is the reason a derivative sign controls the shape of a graph. If $f'(x)>0$ on an interval, take any two points in it; the theorem gives a $c$ between them with $f'(c)$ equal to the average rate, and that average is positive, so the function increases. The same argument gives $f'(x)<0$ implying decrease and $f'(x)=0$ everywhere implying constant. Unit 5.5 uses this result constantly, and it is a theorem consequence rather than a definition.

Common Misconception Check

Myth 1: "The theorem tells you the value of $c$ ." Truth: It tells you one exists. Any specific $c$ has to be found by solving $f'(c)$ equal to the average, and the theorem does not help with that.
Myth 2: "Continuity on the closed interval is enough." Truth: Differentiability on the open interval is a separate condition, and $\lvert x\rvert$ on $[-1,1]$ is the standard counterexample that shows why.
Myth 3: "If the average rate is zero, the function is constant." Truth: It only means the endpoints have the same value, and the derivative is zero somewhere inside. A parabola on a symmetric interval does exactly this without being constant.

Topic-Specific Study Method

Name It, Check It, Then Quote It

Borrowed idea: Name the theorem first. The name brings both hypotheses. Those hypotheses are what a reader checks. Put the interval beside the $c$ . A $c$ without an interval has no justification. The table already hands you the average rate. Spend your time on the condition sentence. So which interval do you name when the table has five rows? The one whose endpoints produce the target number.

What the stem gives you	What you compute	What you write
a table of values	average rate between two rows	the two hypotheses, then the conclusion
a formula	$f'(c)$ set equal to the average	the interval containing $c$
a graph with a corner	nothing, the theorem fails	the reason differentiability fails
a rate function and a target value	average rate over the named interval	existence of $t$ with $f'(t)$ equal to that target

① Write the hypotheses in the order the theorem states them, continuous first, then differentiable. The order signals that you know which property lives on which interval.

② Put the interval inside parentheses right after $c$ . A sentence ending at " $c$ exists" invites the reader to wonder where.

③ When the question asks "must there exist", the answer is yes or no plus a reason. It is never a number, and a number there costs the justification point.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x)=x^{2}$ on $[1,3]$ . Find the value of $c$ guaranteed by the Mean Value Theorem.

Standard Execution:

Check. $f$ is a polynomial, so it is continuous on $[1,3]$ and differentiable on $(1,3)$ .

Average rate. $\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4$ .

Set equal. $f'(x)=2x$ , so $2c=4$ and $c=2$ .

Confirm the interval. $c=2$ lies in $(1,3)$ , so the theorem's promise is met at that point.

Exam Variation (Context / Parameter Shift): A twice-differentiable function $H$ has $H(3)=2$ and $H(5)=6$ . Explain why there must be a time $t$ with $3<t<5$ such that $H'(t)=2$ .

Solution: The average rate on $[3,5]$ is $\frac{6-2}{5-3}=2$ . Because $H$ is twice differentiable it is continuous on $[3,5]$ and differentiable on $(3,5)$ , so the Mean Value Theorem gives a $c$ in $(3,5)$ with $H'(c)=2$ . The computation of the average is the whole argument; the theorem supplies the existence.

Common Traps & Failure Signals:

The unnamed theorem (Signal: "there exists a $c$ " with no justification): the conclusion was quoted without the hypotheses that license it.
The invented location (Signal: a specific $c$ offered as the answer to a "must there exist" question): the existence claim was mistaken for a computation.
The wrong interval (Signal: $c$ outside $(a,b)$ ): the average was taken over one interval and the conclusion attached to another.

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

No Name: You argued from the picture alone.
Hypothesis Skipped: You quoted the conclusion on an interval where the derivative failed somewhere inside, which is exactly the case the theorem excludes and the case the graders were taught to test.
Average Misplaced: You divided by the wrong width, so the target value never matched the derivative and the answer looked like a computational slip rather than a conceptual one.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Three released BC sets make the theorem the reason a value must exist, and in two of them the function is supplied as a table rather than a formula
10-Year Total Count: 3 of 9 Exams · 3 FRQ Parts; Counted where the scored step is the theorem's guarantee, not merely an average rate computed for its own sake
FRQ Years Appeared: 2018 (Q4b), 2021 (Q4d), 2025 (Q1B)

Practice Examples

MCQ 1

A Guaranteed Point

Let $f(x)=x^{3}$ on the interval $[0,2]$ . What value of $c$ satisfies the conclusion of the Mean Value Theorem?

MCQ 2

Which Interval Breaks It

On which interval does the Mean Value Theorem fail to apply to $f(x)=\lvert x-2\rvert$ ?

MCQ 3

Reading a Conclusion

A differentiable function $g$ has $g(1)=4$ and $g(5)=4$ . Which statement must be true?

FRQ 1 • One Interval, Then Two

The function $f$ is twice differentiable on $[0,12]$ . Selected values are given in the table.

$x$	$0$	$3$	$6$	$9$	$12$
$f(x)$	$5$	$11$	$17$	$14$	$20$

(a) Find the average rate of change of $f$ on $[0,6]$ .

(b) Explain why there must be a value $c$ in $(0,6)$ with $f'(c)$ equal to the value from part (a).

(c) Must there be a value $d$ in $(6,12)$ with $f'(d)=2$ ? Justify your answer.

(d) The graph of $f'$ is known to be continuous on $[0,12]$ . Explain why there must be a value $e$ in $(6,12)$ with $f'(e)=1$ .

Cheatsheet

Mean Value Theorem

	What to write	Common error
hypotheses	continuous on $[a,b]$ , differentiable on $(a,b)$	claiming only continuity
average rate	$\dfrac{f(b)-f(a)}{b-a}$	dividing by the wrong width
conclusion	some $c$ in $(a,b)$ with $f'(c)$ equal to it	naming a specific $c$
guarantee	existence only	promising uniqueness or location
failure mode	a corner or a vertical asymptote inside	blaming the endpoints

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Unit 5MVT Statement

Concept 4.231Built for May 2027 AP Exam

MVT Statement

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: A question that asks you to show some rate must have occurred, phrased as "is there a time $t$ " or "must there exist a value $c$ ", usually with a table of values instead of a formula.
What you’ll do: Compute the average rate over the stated interval, state that the function is continuous and differentiable, then invoke the theorem to produce a $c$ with $f'(c)$ equal to that average.
Memory hook: Average in, instant out. The average rate is the input, and the theorem hands back an instant where that rate happens.
Use it on the exam: For MCQs: The distractors usually offer a specific value of $c$ , which the theorem cannot supply, or they let the theorem apply on an interval where differentiability fails. For FRQs: On 2021 Q4(d) the guideline gives a point for the average rate and a separate point for the correct conclusion about whether the theorem guarantees such a $c$ , and a response that only computes the average loses half of it.

Concept Review

The Statement, Split into Hypotheses and Conclusion

Let $f$ satisfy two conditions: continuous on $[a,b]$ , differentiable on $(a,b)$ . Then there is at least one $c$ in $(a,b)$ with

$f'(c) = \frac{f(b)-f(a)}{b-a}$

A concave down curve with its secant line between two endpoints and a parallel tangent line touching the curve at an interior point c

What the Theorem Does and Does Not Promise

Reading It as an Average-to-Instant Translation

$\text{average rate over }[a,b] \;=\; \text{some } f'(c)$

That is why the exam asks the question with a table. The table supplies the average, the derivative supplies the instant, and the theorem supplies the guarantee that the two meet.

The Two Ways the Hypotheses Fail

A corner. For $f(x)=\lvert x\rvert$ on $[-1,1]$ , the average rate is $0$ , but no tangent has slope $0$ , because the derivative does not exist at $0$ . Continuity alone is not enough.
A vertical asymptote. For $f(x)=1/x$ on $[-1,1]$ , the function is not even continuous on the closed interval, so the theorem is silent and the average rate is meaningless.

Both failures are drawn from the same idea. The hypotheses are exactly the conditions that rule out the two ways a curve can refuse to have a parallel tangent.

The Consequences You Will Use Later

Common Misconception Check

Myth 1: "The theorem tells you the value of $c$ ." Truth: It tells you one exists. Any specific $c$ has to be found by solving $f'(c)$ equal to the average, and the theorem does not help with that.
Myth 2: "Continuity on the closed interval is enough." Truth: Differentiability on the open interval is a separate condition, and $\lvert x\rvert$ on $[-1,1]$ is the standard counterexample that shows why.
Myth 3: "If the average rate is zero, the function is constant." Truth: It only means the endpoints have the same value, and the derivative is zero somewhere inside. A parabola on a symmetric interval does exactly this without being constant.

Topic-Specific Study Method

Name It, Check It, Then Quote It

What the stem gives you	What you compute	What you write
a table of values	average rate between two rows	the two hypotheses, then the conclusion
a formula	$f'(c)$ set equal to the average	the interval containing $c$
a graph with a corner	nothing, the theorem fails	the reason differentiability fails
a rate function and a target value	average rate over the named interval	existence of $t$ with $f'(t)$ equal to that target

① Write the hypotheses in the order the theorem states them, continuous first, then differentiable. The order signals that you know which property lives on which interval.

② Put the interval inside parentheses right after $c$ . A sentence ending at " $c$ exists" invites the reader to wonder where.

③ When the question asks "must there exist", the answer is yes or no plus a reason. It is never a number, and a number there costs the justification point.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x)=x^{2}$ on $[1,3]$ . Find the value of $c$ guaranteed by the Mean Value Theorem.

Standard Execution:

Check. $f$ is a polynomial, so it is continuous on $[1,3]$ and differentiable on $(1,3)$ .

Average rate. $\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4$ .

Set equal. $f'(x)=2x$ , so $2c=4$ and $c=2$ .

Confirm the interval. $c=2$ lies in $(1,3)$ , so the theorem's promise is met at that point.

Exam Variation (Context / Parameter Shift): A twice-differentiable function $H$ has $H(3)=2$ and $H(5)=6$ . Explain why there must be a time $t$ with $3<t<5$ such that $H'(t)=2$ .

Solution: The average rate on $[3,5]$ is $\frac{6-2}{5-3}=2$ . Because $H$ is twice differentiable it is continuous on $[3,5]$ and differentiable on $(3,5)$ , so the Mean Value Theorem gives a $c$ in $(3,5)$ with $H'(c)=2$ . The computation of the average is the whole argument; the theorem supplies the existence.

Common Traps & Failure Signals:

The unnamed theorem (Signal: "there exists a $c$ " with no justification): the conclusion was quoted without the hypotheses that license it.
The invented location (Signal: a specific $c$ offered as the answer to a "must there exist" question): the existence claim was mistaken for a computation.
The wrong interval (Signal: $c$ outside $(a,b)$ ): the average was taken over one interval and the conclusion attached to another.

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

No Name: You argued from the picture alone.
Hypothesis Skipped: You quoted the conclusion on an interval where the derivative failed somewhere inside, which is exactly the case the theorem excludes and the case the graders were taught to test.
Average Misplaced: You divided by the wrong width, so the target value never matched the derivative and the answer looked like a computational slip rather than a conceptual one.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Three released BC sets make the theorem the reason a value must exist, and in two of them the function is supplied as a table rather than a formula
10-Year Total Count: 3 of 9 Exams · 3 FRQ Parts; Counted where the scored step is the theorem's guarantee, not merely an average rate computed for its own sake
FRQ Years Appeared: 2018 (Q4b), 2021 (Q4d), 2025 (Q1B)

Practice Examples

MCQ 1

A Guaranteed Point

Let $f(x)=x^{3}$ on the interval $[0,2]$ . What value of $c$ satisfies the conclusion of the Mean Value Theorem?

MCQ 2

Which Interval Breaks It

On which interval does the Mean Value Theorem fail to apply to $f(x)=\lvert x-2\rvert$ ?

MCQ 3

Reading a Conclusion

A differentiable function $g$ has $g(1)=4$ and $g(5)=4$ . Which statement must be true?

FRQ 1 • One Interval, Then Two

The function $f$ is twice differentiable on $[0,12]$ . Selected values are given in the table.

$x$	$0$	$3$	$6$	$9$	$12$
$f(x)$	$5$	$11$	$17$	$14$	$20$

(a) Find the average rate of change of $f$ on $[0,6]$ .

(b) Explain why there must be a value $c$ in $(0,6)$ with $f'(c)$ equal to the value from part (a).

(c) Must there be a value $d$ in $(6,12)$ with $f'(d)=2$ ? Justify your answer.

(d) The graph of $f'$ is known to be continuous on $[0,12]$ . Explain why there must be a value $e$ in $(6,12)$ with $f'(e)=1$ .

Cheatsheet

Mean Value Theorem

	What to write	Common error
hypotheses	continuous on $[a,b]$ , differentiable on $(a,b)$	claiming only continuity
average rate	$\dfrac{f(b)-f(a)}{b-a}$	dividing by the wrong width
conclusion	some $c$ in $(a,b)$ with $f'(c)$ equal to it	naming a specific $c$
guarantee	existence only	promising uniqueness or location
failure mode	a corner or a vertical asymptote inside	blaming the endpoints

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Unit 5MVT Statement

Concept 4.231Built for May 2027 AP Exam

MVT Statement

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: A question that asks you to show some rate must have occurred, phrased as "is there a time $t$ " or "must there exist a value $c$ ", usually with a table of values instead of a formula.
What you’ll do: Compute the average rate over the stated interval, state that the function is continuous and differentiable, then invoke the theorem to produce a $c$ with $f'(c)$ equal to that average.
Memory hook: Average in, instant out. The average rate is the input, and the theorem hands back an instant where that rate happens.
Use it on the exam: For MCQs: The distractors usually offer a specific value of $c$ , which the theorem cannot supply, or they let the theorem apply on an interval where differentiability fails. For FRQs: On 2021 Q4(d) the guideline gives a point for the average rate and a separate point for the correct conclusion about whether the theorem guarantees such a $c$ , and a response that only computes the average loses half of it.

Concept Review

The Statement, Split into Hypotheses and Conclusion

Let $f$ satisfy two conditions: continuous on $[a,b]$ , differentiable on $(a,b)$ . Then there is at least one $c$ in $(a,b)$ with

$f'(c) = \frac{f(b)-f(a)}{b-a}$

A concave down curve with its secant line between two endpoints and a parallel tangent line touching the curve at an interior point c

What the Theorem Does and Does Not Promise

Reading It as an Average-to-Instant Translation

$\text{average rate over }[a,b] \;=\; \text{some } f'(c)$

That is why the exam asks the question with a table. The table supplies the average, the derivative supplies the instant, and the theorem supplies the guarantee that the two meet.

The Two Ways the Hypotheses Fail

A corner. For $f(x)=\lvert x\rvert$ on $[-1,1]$ , the average rate is $0$ , but no tangent has slope $0$ , because the derivative does not exist at $0$ . Continuity alone is not enough.
A vertical asymptote. For $f(x)=1/x$ on $[-1,1]$ , the function is not even continuous on the closed interval, so the theorem is silent and the average rate is meaningless.

Both failures are drawn from the same idea. The hypotheses are exactly the conditions that rule out the two ways a curve can refuse to have a parallel tangent.

The Consequences You Will Use Later

Common Misconception Check

Myth 1: "The theorem tells you the value of $c$ ." Truth: It tells you one exists. Any specific $c$ has to be found by solving $f'(c)$ equal to the average, and the theorem does not help with that.
Myth 2: "Continuity on the closed interval is enough." Truth: Differentiability on the open interval is a separate condition, and $\lvert x\rvert$ on $[-1,1]$ is the standard counterexample that shows why.
Myth 3: "If the average rate is zero, the function is constant." Truth: It only means the endpoints have the same value, and the derivative is zero somewhere inside. A parabola on a symmetric interval does exactly this without being constant.

Topic-Specific Study Method

Name It, Check It, Then Quote It

What the stem gives you	What you compute	What you write
a table of values	average rate between two rows	the two hypotheses, then the conclusion
a formula	$f'(c)$ set equal to the average	the interval containing $c$
a graph with a corner	nothing, the theorem fails	the reason differentiability fails
a rate function and a target value	average rate over the named interval	existence of $t$ with $f'(t)$ equal to that target

① Write the hypotheses in the order the theorem states them, continuous first, then differentiable. The order signals that you know which property lives on which interval.

② Put the interval inside parentheses right after $c$ . A sentence ending at " $c$ exists" invites the reader to wonder where.

③ When the question asks "must there exist", the answer is yes or no plus a reason. It is never a number, and a number there costs the justification point.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x)=x^{2}$ on $[1,3]$ . Find the value of $c$ guaranteed by the Mean Value Theorem.

Standard Execution:

Check. $f$ is a polynomial, so it is continuous on $[1,3]$ and differentiable on $(1,3)$ .

Average rate. $\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4$ .

Set equal. $f'(x)=2x$ , so $2c=4$ and $c=2$ .

Confirm the interval. $c=2$ lies in $(1,3)$ , so the theorem's promise is met at that point.

Exam Variation (Context / Parameter Shift): A twice-differentiable function $H$ has $H(3)=2$ and $H(5)=6$ . Explain why there must be a time $t$ with $3<t<5$ such that $H'(t)=2$ .

Solution: The average rate on $[3,5]$ is $\frac{6-2}{5-3}=2$ . Because $H$ is twice differentiable it is continuous on $[3,5]$ and differentiable on $(3,5)$ , so the Mean Value Theorem gives a $c$ in $(3,5)$ with $H'(c)=2$ . The computation of the average is the whole argument; the theorem supplies the existence.

Common Traps & Failure Signals:

The unnamed theorem (Signal: "there exists a $c$ " with no justification): the conclusion was quoted without the hypotheses that license it.
The invented location (Signal: a specific $c$ offered as the answer to a "must there exist" question): the existence claim was mistaken for a computation.
The wrong interval (Signal: $c$ outside $(a,b)$ ): the average was taken over one interval and the conclusion attached to another.

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

No Name: You argued from the picture alone.
Hypothesis Skipped: You quoted the conclusion on an interval where the derivative failed somewhere inside, which is exactly the case the theorem excludes and the case the graders were taught to test.
Average Misplaced: You divided by the wrong width, so the target value never matched the derivative and the answer looked like a computational slip rather than a conceptual one.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Three released BC sets make the theorem the reason a value must exist, and in two of them the function is supplied as a table rather than a formula
10-Year Total Count: 3 of 9 Exams · 3 FRQ Parts; Counted where the scored step is the theorem's guarantee, not merely an average rate computed for its own sake
FRQ Years Appeared: 2018 (Q4b), 2021 (Q4d), 2025 (Q1B)

Practice Examples

MCQ 1

A Guaranteed Point

Let $f(x)=x^{3}$ on the interval $[0,2]$ . What value of $c$ satisfies the conclusion of the Mean Value Theorem?

MCQ 2

Which Interval Breaks It

On which interval does the Mean Value Theorem fail to apply to $f(x)=\lvert x-2\rvert$ ?

MCQ 3

Reading a Conclusion

A differentiable function $g$ has $g(1)=4$ and $g(5)=4$ . Which statement must be true?

FRQ 1 • One Interval, Then Two

The function $f$ is twice differentiable on $[0,12]$ . Selected values are given in the table.

$x$	$0$	$3$	$6$	$9$	$12$
$f(x)$	$5$	$11$	$17$	$14$	$20$

(a) Find the average rate of change of $f$ on $[0,6]$ .

(b) Explain why there must be a value $c$ in $(0,6)$ with $f'(c)$ equal to the value from part (a).

(c) Must there be a value $d$ in $(6,12)$ with $f'(d)=2$ ? Justify your answer.

(d) The graph of $f'$ is known to be continuous on $[0,12]$ . Explain why there must be a value $e$ in $(6,12)$ with $f'(e)=1$ .

Cheatsheet

Mean Value Theorem

	What to write	Common error
hypotheses	continuous on $[a,b]$ , differentiable on $(a,b)$	claiming only continuity
average rate	$\dfrac{f(b)-f(a)}{b-a}$	dividing by the wrong width
conclusion	some $c$ in $(a,b)$ with $f'(c)$ equal to it	naming a specific $c$
guarantee	existence only	promising uniqueness or location
failure mode	a corner or a vertical asymptote inside	blaming the endpoints

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