Mean Value Theorem (MVT)
The MVT is a core existence theorem: it guarantees there’s at least one point where the instantaneous rate of change equals the average rate of change.
If f(x) is:
- Continuous on [a, b]
- Differentiable on (a, b)
Then there exists at least one c in (a, b) such that:
(Slope of Tangent = Slope of Secant)
Question: Explain why there must be a value c where f'(c) = 2.
Answer: "Since f(x) is continuous on [0, 4] and differentiable on (0, 4), by the Mean Value Theorem, there exists a c such that f'(c) = (f(4)-f(0))/(4-0) = 2."
A function f is continuous on [1,5] and differentiable on (1,5). If f(1)=3 and f(5)=-1, which statement must be true?
Extreme Value Theorem (EVT) + Critical Points
EVT is the existence guarantee for absolute extrema — but only when the conditions match.
If f is continuous on a closed interval [a,b], then f attains both an absolute maximum and an absolute minimum on [a,b].
Big AP trap: EVT does not tell you where the extrema are — only that they exist.
- Absolute (Global) Max/Min: highest/lowest output on the entire interval.
- Relative (Local) Max/Min: highest/lowest output compared to nearby x-values.
- Critical Point: where f'(x)=0 or f'(x) DNE (and x is in the domain).
- Closed interval? [a,b] ✅
- Continuous on the entire interval? ✅
- If yes → absolute max/min exist. If no → can fail.
Intervals of Increase/Decrease (Using f')
The sign of f'(x) tells you whether f is going uphill or downhill. AP expects you to translate between sign charts, graphs, and words quickly.
f'(x) > 0
f'(x) < 0
f'(x) = 0 or DNE
“Because f'(x) > 0 on (a,b), f is increasing on (a,b). Because f'(x) < 0 on (b,c), f is decreasing on (b,c).”
First Derivative Test (Relative/Local Extrema)
This test uses a sign change in f' to classify a critical point as a local max/min.
- Local Max: f' changes from + to −.
- Local Min: f' changes from − to +.
- No Extrema: f' does not change sign.
Critical point ≠ automatically an extremum. You must show a sign change (or use another valid justification).
The Candidates Test (Absolute/Global Extrema)
When finding the Absolute Max/Min on a closed interval [a, b], you must check critical points and endpoints.
- Find critical points (solve f'(x)=0 or where f'(x) DNE).
- Keep only those within [a,b].
- Evaluate f at endpoints and valid critical points.
- Largest output → absolute max; smallest output → absolute min.
A correct Candidates Test solution usually needs a table (or clear list) of x-values checked and the corresponding f(x) values.
Concavity (Using f'')
Concavity describes how the graph “bends.” The second derivative tells whether slopes are increasing or decreasing.
Concave Up (∪)
f''(x) > 0
"Smiley Face"
Slopes are increasing.
Concave Down (∩)
f''(x) < 0
"Frown Face"
Slopes are decreasing.
A point where the graph changes concavity. This occurs when f'' changes sign.
Warning: f''(c)=0 does not guarantee a POI (example: y=x^4 at x=0).
Second Derivative Test (Classifying Critical Points)
The second derivative test is a quick classification tool, but only works when f'(c)=0 and f''(c) exists.
- If f'(c)=0 and f''(c)>0 → local minimum.
- If f'(c)=0 and f''(c)<0 → local maximum.
- If f''(c)=0 → inconclusive (use First Derivative Test or sign chart).
“Since f'(2)=0 and f''(2)<0, by the Second Derivative Test, f has a relative maximum at x=2.”
Sketching Graphs of f and f'
AP loves “graph translation” questions: given a graph of f, sketch f' (or vice versa). Your job is to track slopes, not y-values.
- Where f has a horizontal tangent → f'(x)=0.
- Where f is increasing → f' is above the x-axis.
- Where f is decreasing → f' is below the x-axis.
- Where f is steep → |f'| is large.
If a graph of f has a local maximum at x=3, what must be true about f' at x=3?
Connecting f, f', and f''
You must be fluent in translating between the three levels of a function — especially from a graph of f' or f''.
| Relationship | What f does | What f' does | What f'' does |
|---|---|---|---|
| Increasing | Going Up | Positive (+) | (No direct link) |
| Decreasing | Going Down | Negative (-) | (No direct link) |
| Concave Up | Bowl Shape (∪) | Increasing ↗ | Positive (+) |
| Concave Down | Hill Shape (∩) | Decreasing ↘ | Negative (-) |
| POI | Change Curve | (Not always special) | Change Sign |
If you’re given a graph of f':
- f increases where f' is above the x-axis.
- f has a local max/min where f' crosses the x-axis with a sign change.
- f is concave up where f' is increasing.
Introduction to Optimization (Setup)
Optimization problems ask for “largest/smallest possible” under a constraint. The hardest part is usually the modeling, not the derivative.
- Primary Equation: the thing you optimize (Area, Volume, Cost, Profit).
- Constraint: a second equation relating variables (given info).
- Substitute: rewrite the primary equation using one variable.
- Differentiate: set derivative to zero to find candidates.
- Verify: justify max/min (test, endpoints, or reasoning).
Solving Optimization Problems (AP Method + FRQ Language)
A full-credit AP optimization solution is usually: define variables → build equations → optimize → justify.
- Let x = __________ and y = __________.
- Constraint: _______________________ (given info)
- Quantity to optimize: Q = _______________________
- Rewrite: Substitute constraint into Q so Q is a function of one variable.
- Differentiate: Find Q'(x), solve Q'(x)=0.
- Justify: Use a test/endpoints/reasoning to confirm max/min.
- Answer in context: include units and a sentence.
If the problem is on a closed interval, many rubrics expect a Candidates Test (critical points + endpoints) for justification.
Exploring Behaviors of Implicit Relations
Implicit relations can have critical points even when they’re not functions. AP often asks you to find where the tangent is horizontal or vertical.
- Horizontal tangent: dy/dx = 0 → numerator = 0 and denominator ≠ 0.
- Vertical tangent: dy/dx undefined → denominator = 0 and numerator ≠ 0.
- Always verify the point is actually on the relation.
Given an implicit relation and dy/dx, you’re often asked: “Find all points where the tangent is horizontal/vertical.”
- Set numerator (or denominator) to 0 depending on the tangent type.
- Solve for x and y, then plug back into the original relation.
- Reject extraneous points not on the curve.
Must-Memorize Checklist (Ace a 5)
This is the Unit 5 “no excuses” list. If you can recite these conditions + write the FRQ sentence frames, you’ll pick up easy points fast.
- MVT: continuous on [a,b], differentiable on (a,b) → ∃ c with f'(c)= avg slope.
- EVT: continuous on [a,b] → absolute max & min exist on [a,b].
- Critical point: f'(x)=0 or DNE (in domain).
- Inc/Dec: f'(x)>0 inc, f'(x)<0 dec.
- 1st Derivative Test:
+→− local max, −→+ local min, no sign change = none. - 2nd Derivative Test: if f'(c)=0 and f''(c)>0 min, f''(c)<0 max, f''(c)=0 inconclusive.
- Concavity: f''>0 up, f''<0 down.
- Inflection: concavity changes → f'' changes sign.
- Implicit tangents: horizontal when dy/dx=0 (num=0, den≠0), vertical when undefined (den=0, num≠0).
“Since f is continuous on [a,b] and differentiable on (a,b), by the Mean Value Theorem there exists c in (a,b) such that f'(c)=\frac{f(b)-f(a)}{b-a}.”
“Because f'(x)>0 on (a,b), f is increasing on (a,b). Because f'(x)<0 on (b,c), f is decreasing on (b,c).”
“Because f''(x)>0 on (a,b), f is concave up on (a,b). Because f''(x) changes sign at x=c, f has a point of inflection at x=c.”
“Let Q be the quantity to optimize. Using the constraint, write Q as a function of one variable, find critical points by solving Q'(x)=0, and verify the maximum/minimum with an appropriate test. Therefore, the optimal value is ______ (units).”
Graph Matching: Given f, Choose the Correct f′
This is a high-frequency AP skill. Focus on slope: where f increases, f′ is above 0; where f decreases, f′ is below 0; where f levels off, f′ hits 0.