Analytical Applications of Differentiation: MVT and Extrema (AP Calculus AB Unit 5)

0.0(0)
Studied by 0 people
0%Unit 5 Mastery
0%Exam Mastery
Build your Mastery score
multiple choiceMultiple Choice
call kaiCall Kai
Supplemental Materials
Card Sorting

1/24

Last updated 3:08 PM on 3/12/26
Name
Mastery
Learn
Test
Matching
Spaced
Call with Kai

No analytics yet

Send a link to your students to track their progress

25 Terms

1
New cards

Mean Value Theorem (MVT)

If f is continuous on [a,b] and differentiable on (a,b), then there exists c in (a,b) such that f'(c) = (f(b)-f(a))/(b-a).

2
New cards

Average rate of change

The slope of the secant line over [a,b]: (f(b)-f(a))/(b-a).

3
New cards

Instantaneous rate of change

The derivative value f'(c), representing the slope of the tangent line at x=c.

4
New cards

Secant line

The line through (a,f(a)) and (b,f(b)); its slope is the average rate of change on [a,b].

5
New cards

Tangent line

The line that touches the curve at a point x=c with slope f'(c).

6
New cards

MVT hypotheses

The conditions required to apply MVT: f must be continuous on the closed interval [a,b] and differentiable on the open interval (a,b).

7
New cards

MVT conclusion (existence of c)

MVT guarantees at least one c in (a,b) where the tangent slope equals the secant slope, but it does not tell you where c is.

8
New cards

Non-uniqueness of c (in MVT)

MVT does not guarantee exactly one c; there may be multiple points in (a,b) satisfying f'(c) = (f(b)-f(a))/(b-a).

9
New cards

Rolle’s Theorem

Special case of MVT: if f is continuous on [a,b], differentiable on (a,b), and f(a)=f(b), then there exists c in (a,b) such that f'(c)=0.

10
New cards

Jump discontinuity (why MVT can fail)

A break in the graph on [a,b] (function not continuous), which can prevent MVT from applying.

11
New cards

Corner/Cusp (why MVT can fail)

A sharp point where f'(x) does not exist in (a,b), so differentiability fails and MVT may not apply.

12
New cards

Extreme Value Theorem (EVT)

If f is continuous on a closed interval [a,b], then f attains an absolute maximum and an absolute minimum on [a,b].

13
New cards

Absolute maximum value

The greatest output f(x) achieves on a domain/interval; occurs at some xmax with f(xmax) ≥ f(x) for all x in the interval.

14
New cards

Absolute minimum value

The least output f(x) achieves on a domain/interval; occurs at some xmin with f(xmin) ≤ f(x) for all x in the interval.

15
New cards

Closed interval requirement (EVT)

EVT requires endpoints included (interval [a,b]); on open intervals, a continuous function may fail to attain max/min.

16
New cards

Continuity requirement (EVT)

EVT requires f be continuous on [a,b]; discontinuities (e.g., vertical asymptotes or undefined points) can prevent absolute extrema from existing.

17
New cards

Relative (local) maximum

A point x=c where f(c) is greater than or equal to nearby values of f(x) in some open interval around c (not necessarily the greatest overall).

18
New cards

Relative (local) minimum

A point x=c where f(c) is less than or equal to nearby values of f(x) in some open interval around c (not necessarily the least overall).

19
New cards

Critical number

A domain value c where f'(c)=0 or where f'(c) does not exist (as long as f(c) exists).

20
New cards

First Derivative Test

Method to classify critical numbers using sign changes of f': + to − gives a relative max; − to + gives a relative min; no sign change gives no local extremum.

21
New cards

Increasing interval (via derivative)

An interval where f'(x) > 0, meaning f increases as x increases.

22
New cards

Decreasing interval (via derivative)

An interval where f'(x) < 0, meaning f decreases as x increases.

23
New cards

Candidates Test (for absolute extrema)

For a continuous f on [a,b], absolute extrema occur at endpoints or interior critical numbers; evaluate f at all candidates and compare values.

24
New cards

Endpoint candidates

The interval endpoints x=a and x=b, which must be included as possible locations of absolute maxima/minima in the Candidates Test.

25
New cards

Interior candidate restriction

Only critical numbers that lie inside (a,b) are candidates for absolute extrema on [a,b]; solutions outside the interval are ignored.

Explore top flashcards

flashcards
Quarter 2 Vocabulary Words
30
Updated 861d ago
0.0(0)
flashcards
human geo unit 3 gradesavers
69
Updated 1219d ago
0.0(0)
flashcards
Memory
42
Updated 874d ago
0.0(0)
flashcards
Unit 7: Natural Selection
35
Updated 1073d ago
0.0(0)
flashcards
Polytomic Ions
29
Updated 1015d ago
0.0(0)
flashcards
Vocab G Unit 6
20
Updated 1215d ago
0.0(0)
flashcards
Antidiabetic Drugs
52
Updated 1201d ago
0.0(0)
flashcards
EM E2: Trauma
87
Updated 325d ago
0.0(0)
flashcards
Quarter 2 Vocabulary Words
30
Updated 861d ago
0.0(0)
flashcards
human geo unit 3 gradesavers
69
Updated 1219d ago
0.0(0)
flashcards
Memory
42
Updated 874d ago
0.0(0)
flashcards
Unit 7: Natural Selection
35
Updated 1073d ago
0.0(0)
flashcards
Polytomic Ions
29
Updated 1015d ago
0.0(0)
flashcards
Vocab G Unit 6
20
Updated 1215d ago
0.0(0)
flashcards
Antidiabetic Drugs
52
Updated 1201d ago
0.0(0)
flashcards
EM E2: Trauma
87
Updated 325d ago
0.0(0)