Why must an instantaneous rate sometimes match an average rate?
You will be able to: Check the Mean Value Theorem hypotheses and find an interior matching rate.
Why must an instantaneous rate sometimes match an average rate?
A cart moves from position 1 meter to position 9 meters between 1 and 3 seconds. Its average velocity is 4 m/s. If its position is continuous and differentiable as required, it must have velocity 4 m/s at some time between them.
A useful starting point: Prerequisite: derivative signs and tangent-line behavior →
Words and symbols before equations
- Closed interval [a,b]
- Includes both endpoints.
- Open interval (a,b)
- Excludes both endpoints.
- Secant slope
- The endpoint change divided by the input change.
- Mean Value Theorem (MVT)
- Under continuity and differentiability hypotheses, an interior derivative equals the secant slope.
What this picture assumes
Original model; numerical readouts are rounded. s=t² meters on [1,3] seconds; continuous on the closed interval and differentiable inside. Fixed secant slope 4 m/s. Tangent control remains strictly inside.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- At c=2 s the teal tangent slope is 4 m/s. Fixed orange secant slope=4 m/s. They match at c=2. Blue curve s=t²; the MVT hypotheses hold.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
The MVT requires f continuous on [a,b] and differentiable on (a,b). Then some c strictly between a and b satisfies f′(c)=(f(b)−f(a))/(b−a). Check both hypotheses before claiming a guarantee.
For s(t)=t² meters on [1,3] seconds, the polynomial is continuous and differentiable everywhere. Its average rate is (9−1)/(3−1)=4 m/s.
Since s′(t)=2t, solve 2c=4 to obtain c=2 s. The orange secant and tangent at c are parallel, but they need not lie on top of one another.
The theorem guarantees at least one c, not necessarily a unique one. It concerns a derivative value, not a function height or an average of endpoint derivatives.
A worked example, step by step
Find the MVT point for f(x)=x² on [2,6].
- The polynomial is continuous on [2,6] and differentiable on (2,6).
- The secant slope is (36−4)/(6−2)=8.
- Solve f′(c)=2c=8 to get c=4.
- Since 2<4<6, this point meets the theorem’s interior requirement.
The matching point must be inside the interval; matching heights or using the midpoint without solving is not the theorem.
Does the theorem promise exactly one matching point?
Compare with an explanation
No; it guarantees at least one. This particular quadratic has exactly one.
Predict. Change one thing. Explain.
Move the candidate tangent point along t² on [1,3]. Find where its slope matches the fixed secant and explain why the theorem applies.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At c=2 s the teal tangent slope is 4 m/s. Fixed orange secant slope=4 m/s. They match at c=2. Blue curve s=t²; the MVT hypotheses hold.
Original model; numerical readouts are rounded. s=t² meters on [1,3] seconds; continuous on the closed interval and differentiable inside. Fixed secant slope 4 m/s. Tangent control remains strictly inside.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the relevant derivative signs, theorem conditions, domain or geometric constraint. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionFor f(x)=x²+2x on [0,4], justify the MVT and find every c it supplies.
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Compare with the answer and four-point rubric
- 1 point: A polynomial is continuous on [0,4] and differentiable on (0,4).
- 1 point: The secant slope is (24−0)/4=6.
- 1 point: f′=2x+2, so 2c+2=6 gives c=2.
- 1 point: The value 2 is interior, and the linear derivative equation has no other solution.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What are the two main hypotheses?
Continuity on the closed interval and differentiability on the open interval.
RECALL 2What matches geometrically?
A tangent slope and the endpoint secant slope.
RECALL 3Is c necessarily unique?
No.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why must an instantaneous rate sometimes match an average rate?
- f′(c)=(f(b)−f(a))/(b−a), a<c<b, with the MVT hypotheses.
- Continuity on [a,b]; differentiability on (a,b).
Remember: The matching point must be inside the interval; matching heights or using the midpoint without solving is not the theorem.
Conditions: Original model; numerical readouts are rounded. s=t² meters on [1,3] seconds; continuous on the closed interval and differentiable inside. Fixed secant slope 4 m/s. Tangent control remains strictly inside.
Refresh Kid · AP Calculus AB Unit 5 · Objectives FUN-1.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.1, FUN-1.B. CED effective Fall 2020, with Fall 2026 clarifications, checked September 17, 2026. Unit 5 has twelve official topics. Focused lesson titles and questions are original Refresh Kid teaching material.
Existence theorems require their stated hypotheses. Critical numbers must belong to the function’s domain; interior local extrema and included endpoints are handled explicitly. Extrema and inflection candidates need justification, not just a zero derivative. Sparse derivative samples do not establish interval-wide signs. Optimization includes the constraint, feasible domain and global comparison. Implicit-curve conclusions specify the branch and distinguish finite slopes from vertical tangents.
The Organic Chemistry Tutor Mean Value Theorem and Optimization Problems video titles, creator and relevant descriptions were checked; full videos were not reviewed. The free optimization video mentions additional paid material, which is not required here. Khan Academy’s unit destination was checked; its JavaScript lesson contents were not fully readable by the research tool. OpenStax Sections 4.3, 4.4, 4.5 and 4.7 were consulted for conceptual cross-checking. No creator scripts, questions, diagrams or artwork were copied. Refresh Kid is not affiliated with or endorsed by these providers.
GitHub’s 3D website collection and its camera-control example informed optional spatial inspection. The open-box model is original geometry using existing self-hosted Three.js with its MIT license. A 12 cm square sheet loses four equal corner squares. The remaining net folds into an open box; the geometry uses the same lengths as the labeled 2D net. Fold angle is a construction view, not an independent design variable; displayed volume refers to the fully upright box. Camera rotation changes no mathematical values. Complete 2D diagrams and explanations remain available without WebGL.
Independent teacher review and observation of students remain pending. Technical checks do not certify mathematical accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus AB questions and scoring guides. The archive spans multiple units; no entire exam question is assigned here.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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