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
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 is HORIZONTAL; denominator = 0 is VERTICAL
Implicit, the second test
the candidate must also SATISFY the original equation
Implicit y″
differentiate y′, then substitute y′ back in