Analytical Applications of Differentiation

Twelve topics that turn the derivative back on the function it came from. The Mean Value Theorem and its hypotheses, the Extreme Value Theorem and critical points of both kinds, increasing and decreasing from the sign of the derivative, the first derivative test and the candidates test, concavity and the second derivative test with its one refusal, sketching a function beside its derivatives, optimization from setup through to a justified answer, and implicit relations, where a candidate has to lie on the curve.

AB exam 15-20%BC exam 10-15%12 topics

Topics

5.1

Using the Mean Value Theorem

Extreme Value Theorem, Global Versus Local Extrema, and Critical Points

Determining Intervals on Which a Function Is Increasing or Decreasing

Using the First Derivative Test to Determine Relative (Local) Extrema

Using the Candidates Test to Determine Absolute (Global) Extrema

Determining Concavity of Functions over Their Domains

Using the Second Derivative Test to Determine Extrema

Sketching Graphs of Functions and Their Derivatives

Connecting a Function, Its First Derivative, and Its Second Derivative

Introduction to Optimization Problems

Solving Optimization Problems

Exploring Behaviors of Implicit Relations

Lesson

Diagnostic

after diagnostic

Key forms For every problem in this unit

Mean Value Theorem, hypotheses

continuous on [a, b] AND differentiable on (a, b)

Mean Value Theorem, conclusion

SOME c in (a, b) has f′(c) = (f(b) − f(a))/(b − a)

Not promised

uniqueness, the midpoint, or anything about f(c)

Rolle

same, plus f(a) = f(b), so f′(c) = 0

Extreme Value Theorem

continuous on a CLOSED, BOUNDED interval

Critical point

c in the domain of f with f′(c) = 0 OR f′(c) undefined

First derivative test

+ to − is a MAX; − to + is a MIN; no change is NEITHER

Second derivative test

f′(c) = 0 and f″(c) > 0 is a MIN; f″(c) < 0 is a MAX

When it fails

f″(c) = 0 is INCONCLUSIVE, not a verdict

Candidates test

interior critical points PLUS both endpoints, compare f

Justification

name the derivative, what it does, and where

Increasing / decreasing

sign of f′, never the sign of f

Concave up

f″ > 0, equivalently f′ INCREASING

Independent

all four pairings of direction with concavity occur

Inflection point

f″ CHANGES SIGN, and the point is in the domain

On a graph of f′

above the axis = rising; RISING = concave up

Chart splitters

zeros, undefined points, and gaps in the domain of f

Optimization setup

objective + constraint → ONE variable, then a domain

Domain

from the situation: positivity, caps, whole counts

Answer the question

x, the other dimension, or the extreme VALUE

Distance problems

minimize d²; same location, easier derivative

Implicit tangents

numerator = 0 (denominator ≠ 0) is HORIZONTAL; denominator = 0 (numerator ≠ 0) is VERTICAL. Both zero: 0/0, examine the point separately. Always confirm the point lies on the curve

Implicit, the second test

the candidate must also SATISFY the original equation

Implicit y″

differentiate y′, then substitute y′ back in

Unit 5 tools

Challenge bank

1 / 60

60 open-ended problems.

Read the question, work it out, then flip the card to compare your reasoning to the worked solution. Mark each card so you can return to the ones that still bite.

Question

Tap card to reveal explanation

Worked solution

Tap card to return to question

Cumulative assessment

Test the unit.

Twenty mixed items drawn from across all 12 topics, with guaranteed misconception-code coverage. Identifies which misconceptions still bite when you cannot see which topic the question came from.

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Check what stuck.

Units 1 through 5, drawn evenly so earlier units get the same share as this one. Twenty questions or a full 45-question section, your choice. Even coverage means this is a retention check rather than a score estimate. Follows the track you chose on the course page.

20 or 45questions

51topics

69codes covered

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