Conditions before computation
For MVT, require continuity on [a,b] and differentiability on (a,b). Endpoint differentiability is not required. Rolle additionally requires equal endpoint values; it guarantees at least one interior derivative zero, not exactly one. A bound on all interior derivatives bounds the total change through MVT. Before L’Hôpital, check the original quotient has a 0/0 or ∞/∞ form; if one application leaves an indeterminate quotient, check the conditions again before another application.
Worked example with a method check
f is continuous on [0,2] and differentiable on (0,2), with −3≤f′(x)≤6. If f(0)=6, what is the smallest value f(2) could have under these constraints? The Mean Value Theorem expresses the total change as an interior derivative times interval length. The smallest allowed change is −3×2=-6. Add the initial value to obtain 0. The line f(x)=6−3x attains this bound.
Compare two methods
The function |x| is continuous on [−1,1] with equal endpoint values, but its corner prevents Rolle’s theorem from applying. In fact its derivative is never zero where defined. In contrast, a differentiable polynomial with four distinct roots forces at least three distinct derivative zeros between neighboring roots. A failed hypothesis removes a guarantee; it does not automatically prove that every possible conclusion is false.
Transfer: solve before reading
A function satisfies f(0)=7 and −2≤f′≤4 on (0,3), with MVT hypotheses satisfied. Bound f(3). Separately, explain why replacing lim x→0 (3+x)/(8+x) by 1 is invalid.
Transfer reasoning and answer
The change lies between −6 and 12, so 1≤f(3)≤19. Both endpoint bounds can be attained by suitable affine functions. The quotient tends directly to 3/8 and is not indeterminate; differentiating numerator and denominator to obtain 1 uses L’Hôpital outside its conditions.
Practice the next representation
Work all four task families linked to this lesson. For the illustrated items, state which numerical labels and relationships you used. For a written extension, compare each line of your solution with its original self-assessment criteria; these criteria are practice feedback, not an official AP score.
f is continuous on [0,2] and differentiable on (0,2), with −3≤f′(x)≤6. If f(0)=6, what is the smallest value f(2) could have under these constraints?
- The Mean Value Theorem expresses the total change as an interior derivative times interval length.
- The smallest allowed change is −3×2=-6.
- Add the initial value to obtain 0. The line f(x)=6−3x attains this bound.
Compare the other answer choices
Replace the negative lower rate by its magnitude.
Use the upper derivative bound instead of the lower one.
Report the minimum change without adding f(0).
Explain your thinking.
Optional written extension. Self-assessed only; it does not change your automatic practice result. These are teaching criteria, not official awarded marks.
(a) State the relationship or theorem that makes this calculation valid, including its relevant conditions. (b) Show the intermediate computation and give the requested result. (c) Explain why one plausible competing method would fail for this problem.
This draft stays on this page only. Use paper if you want to keep your working.
Try it, then explain it.
Complete a short set without looking at the example. After each answer, explain which operation preserved the relationship and why the result makes sense.
Sources & authorship
Original Slope explanations and examples. References support the concepts; they are not the source of copied questions. Independent educator review is pending.
AP Calculus AB and BC Course and Exam DescriptionCalculus Volume 1: Fundamental Theorem of Calculus