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AP Calculus AB/BC Unit 5 Visual ReviewA topic-by-topic visual walkthrough of Unit 5: Analytical Applications of Differentiation — the MVT, extrema, the first/second derivative tests, concavity, and optimization.
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TOPIC 5.1
The Mean Value Theorem (MVT)
If f is continuous on [a,b] and differentiable on (a,b),
then f′(c) = [f(b) − f(a)] / (b − a) for some c
a tangent somewhere is parallel to the secant
What it guarantees
At least one point where the instantaneous
rate equals the AVERAGE rate.
Rolle's Theorem is the special case
f(a)=f(b) → f′(c)=0.
Example
Drive 120 mi in 2 h → avg 60 mph.
MVT guarantees you hit EXACTLY 60 mph
at some instant.
Check the hypotheses before applying.
MVT : some tangent slope equals the average rate [f(b)−f(a)]/(b−a).The Review Hub · AP Calculus AB/BC Unit 5
TOPIC 5.2
Extreme Value Theorem & Critical Points
Critical points
A CRITICAL POINT of f is where
f′(x) = 0 or f′(x) DNE
All local extrema occur at critical points
— but not every critical point is an
extremum (e.g. x³ at 0).
Extreme Value Theorem
If f is CONTINUOUS on a CLOSED interval
[a, b], it attains an absolute MAX and an
absolute MIN on that interval.
LOCAL extremum = highest/lowest nearby;
GLOBAL = over the whole interval.
Extrema live at critical points or endpoints
To find absolute extrema on [a, b], check critical points AND the two endpoints (Candidates Test, 5.5).
Critical points: f′ = 0 or DNE ; a continuous f on [a,b] has an abs max & min.
The Review Hub · AP Calculus AB/BC Unit 5
TOPIC 5.3
Increasing & Decreasing Intervals
f′(x) > 0 → f increasing f′(x) < 0 → f decreasing
Sign chart of f′
1. Find critical points (f′=0 or DNE).
2. Test a point in each interval.
3. Record the sign of f′ on each.
f′: + + + | − − − | + + +
The sign tells you increasing/decreasing.
Worked example
f(x)=x³−3x → f′=3x²−3
f′=0 at x=±1.
Increasing on (−∞,−1) & (1,∞);
decreasing on (−1, 1).
Sign changes mark local extrema (5.4).
The sign of f′ gives increasing (+) / decreasing (−) intervals.
The Review Hub · AP Calculus AB/BC Unit 5
TOPIC 5.4
The First Derivative Test
at a critical point, watch the SIGN CHANGE of f′
+ → − gives a local MAX; − → + gives a local MIN
Local MAX
f′ changes from POSITIVE to NEGATIVE.
f′: + + + 0 − − −
The function rises, then falls → a peak.
No sign change → not an extremum.
Local MIN
f′ changes from NEGATIVE to POSITIVE.
f′: − − − 0 + + +
The function falls, then rises → a valley.
Justify extrema by citing the sign change.
f′: + → − is a max, − → + is a min at a critical point.The Review Hub · AP Calculus AB/BC Unit 5
TOPIC 5.5
The Candidates Test
Absolute extrema on [a, b]
The absolute max/min must occur at a
CRITICAL POINT or an ENDPOINT.
1. List all candidates.
2. Evaluate f at each.
3. Largest = max, smallest = min.
Worked example
f(x)=x³−3x on [0, 2]
crit pt in interval: x=1.
f(0)=0, f(1)=−2, f(2)=2
Abs max = 2 (at x=2), abs min = −2
(at x=1).
Compare y-values, not slopes
No sign analysis needed — just evaluate f at every candidate and pick the biggest and smallest outputs.
Absolute extrema: evaluate f at critical points AND endpoints , compare.
The Review Hub · AP Calculus AB/BC Unit 5
TOPIC 5.6
Determining Concavity
f″(x) > 0 → concave UP f″(x) < 0 → concave DOWN
Concave up (holds water)
f″ > 0: slopes increasing, curve bends up.
Inflection points
Where concavity CHANGES (f″ changes
sign).
f″=0 or DNE is a
candidate; verify a sign change.
f″>0 concave up, f″<0 concave down ; inflection where f″ changes sign.The Review Hub · AP Calculus AB/BC Unit 5
TOPIC 5.7
The Second Derivative Test
at a critical point where f′(c) = 0:
f″(c) > 0 → local MIN; f″(c) < 0 → local MAX
Concavity decides
A critical point on a concave-UP curve is
a bottom → minimum.
A critical point on a concave-DOWN curve
is a top → maximum.
f′(2)=0, f″(2)=6>0 → min at x=2
When it's inconclusive
If f″(c) = 0, the test gives NO conclusion.
Fall back on the FIRST derivative test
(check the sign change of f′).
e.g. x⁴ at 0: f″(0)=0 but it's a min
f″(c)>0 → min, f″(c)<0 → max ; if f″(c)=0 use the first derivative test.The Review Hub · AP Calculus AB/BC Unit 5
TOPIC 5.8
Sketching f and Its Derivatives
Translate features between f and f′
Where f has a LOCAL EXTREMUM, f′ has a ZERO (crosses the x-axis).
Where f has an INFLECTION point, f′ has a local extremum and f″ has a zero.
f → f′
f increasing → f′ above the axis (+).
f decreasing → f′ below the axis (−).
f steepest → f′ at its extreme value.
The slope of f becomes the HEIGHT of f′.
f′ → f (reverse)
f′ > 0 → f is rising.
f′ crosses + → − → f has a max.
f′ increasing → f concave up.
Read one graph to describe the other.
The slope of f = the height of f′ ; extrema of f are zeros of f′.
The Review Hub · AP Calculus AB/BC Unit 5
TOPIC 5.9
Connecting f, f′, and f″
Feature of f f′ f″
increasing f′ > 0 —
local max / min f′ = 0, sign change f″ ≷ 0 (2nd deriv test)
concave up f′ increasing f″ > 0
inflection point f′ local extremum f″ = 0, sign change
One family of information
Given any one of f, f′, or f″, you can describe the others' behavior — a core skill for graph and FRQ questions.
f′ gives increasing/extrema; f″ gives concavity/inflection .The Review Hub · AP Calculus AB/BC Unit 5
TOPIC 5.10
Introduction to Optimization
Maximize or minimize a quantity
Optimization finds the best value — the
largest area, least cost, shortest distance.
Write the quantity as a function of ONE
variable, then find its extremum.
Use a CONSTRAINT to eliminate extra variables.
The idea
The optimum occurs at a CRITICAL POINT
of the objective function.
set (objective)′ = 0, solve
Then confirm it's a max or min with a
derivative test.
Reduce to one variable using the constraint
e.g. fixed perimeter P=2ℓ+2w lets you write area A(ℓ) in ℓ alone, then optimize A(ℓ).
Optimize by writing the quantity in one variable, then setting its derivative = 0 .
The Review Hub · AP Calculus AB/BC Unit 5
TOPIC 5.11
Solving Optimization Problems
The procedure
1. Define variables; draw a picture.
2. Write the OBJECTIVE + the CONSTRAINT.
3. Reduce to ONE variable.
4. Differentiate, set = 0, solve.
5. Confirm min/max & check the domain.
Answer the question that was asked, with units.
Fence example
Maximize area with 100 ft of fence:
2ℓ + 2w = 100 → w = 50 − ℓ
A(ℓ) = ℓ(50 − ℓ)
A′ = 50 − 2ℓ = 0 → ℓ = 25
Max area = 25·25 = 625 ft² (a square).
A″ = −2 < 0 confirms a maximum.
Objective + constraint → one variable, differentiate, solve, confirm .
The Review Hub · AP Calculus AB/BC Unit 5
TOPIC 5.12
Behaviors of Implicit Relations
Extrema on implicit curves
Use implicit differentiation to get dy/dx,
then analyze it like any derivative.
HORIZONTAL tangent where dy/dx = 0
(numerator = 0).
VERTICAL tangent where dy/dx is undefined.
Second derivative implicitly
Differentiate dy/dx AGAIN (implicitly) to
get d²y/dx² for concavity.
Substitute the known dy/dx expression
where it appears.
Circle x²+y²=25: dy/dx = −x/y.
Same analysis, curve not solved for y
All the tangent-line, extremum, and concavity tools apply — you just carry both x and y through the work.
On implicit curves, dy/dx=0 → horizontal tangent ; undefined → vertical tangent.
The Review Hub · AP Calculus AB/BC Unit 5
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How to use the visual review
Spend 30 seconds per slide before clicking next. Look at the diagram, then ask yourself: "Could I build this sign chart or solve this problem from memory?"
Use the fullscreen button () on desktop for the best experience. Use arrow keys to navigate. Tap "Show all slides" to jump around.
This is great for review the night before the exam — fast, visual, and covers every idea you need to recognize in Unit 5.