Analytical Applications of Differentiation
~4 weeks · October into November · 10–15%You can compute derivatives. Now: what do they tell you? This unit is the bridge from local information (the slope at a point) to global conclusions (the function is increasing on this whole interval).
That bridge is one theorem, and it's the one every student writes off as filler.
The Mean Value Theorem
then there is some c in (a,b) with f′(c) = [f(b) − f(a)]/(b − a).
Translation: at some instant, your instantaneous rate equalled your average rate. Drive 120 miles in two hours and at some moment the speedometer read exactly 60.
Students think MVT is a curiosity. It is the only thing licensing every inference from a derivative back to the function. Each of these claims is a corollary of MVT and is false without it:
- f′ > 0 on an interval ⟹ f is increasing there. Proof: take any two points p < q in the interval. MVT gives a c with f(q) − f(p) = f′(c)(q − p). Both factors are positive, so f(q) > f(p). Done.
- f′ = 0 everywhere ⟹ f is constant. Same argument: f(q) − f(p) = 0·(q−p) = 0 for every pair.
- Two functions with the same derivative differ by a constant. Apply the previous fact to their difference.
That last one is why the +C exists. In Unit 6 you'll write ∫f dx = F(x) + C and the C will be presented as a rule to remember. It isn't a rule — it's this theorem. And the fact that antiderivatives are unique up to a constant is precisely what makes the whole integral-as-antiderivative program coherent. Without MVT, the Fundamental Theorem doesn't work.
Where MVT itself comes from
The dependency chain, bottom to top: completeness of the reals → EVT → Rolle → MVT → everything in this unit. BC uses all of it and proves none of it. Worth knowing the chain exists, because "why is that true?" bottoms out somewhere real.
Show f(x) = x³ − x satisfies MVT on [0, 2] and find all valid c.
| f is a polynomial → continuous on [0,2], differentiable on (0,2) | state the hypotheses; this is a scored step |
| f(0) = 0, f(2) = 8 − 2 = 6 | |
| average slope = (6 − 0)/(2 − 0) = 3 | |
| f′(x) = 3x² − 1, set equal to 3 | MVT says some c does this |
| 3c² = 4, c = ±2/√3 | |
| c = 2/√3 ≈ 1.155 (reject the negative — not in (0,2)) | always check c lands in the open interval |
Applying MVT to a function with a corner or a discontinuity in the interval is invalid and the conclusion can be false. f(x) = |x| on [−1,1] has average slope 0, but f′ is never 0 — no contradiction, because f isn't differentiable at 0 and the theorem never applied. On free response, name continuity and differentiability explicitly before using it.
Make the endpoints equal and the Mean Value Theorem has to collapse into Rolle's. Take f(x) = x² − 4x on [0, 4]: f(0) = f(4) = 0, so the guaranteed c must have f′(c) = 0. And it does — c = 2.
Any statement of the MVT you write down that does not reduce to that under equal endpoints is misremembered.
Critical points and extrema
Fermat's Theorem: if f has a local extremum at an interior point c and f′(c) exists, then f′(c) = 0.
Why: suppose f′(c) > 0 at a local max. Then f is increasing through c, so points just to the right are higher — contradicting that c is a max. Same argument with the sign flipped rules out f′(c) < 0. The only survivor is 0.
The converse is false and this is the most common misunderstanding in the unit. f(x) = x³ has f′(0) = 0 but no extremum there — the curve flattens and keeps going. A critical point is a candidate, not a conclusion. You must always test.
First derivative test
At a critical point c, examine the sign of f′ on either side:
- + to − → local maximum (rising then falling)
- − to + → local minimum (falling then rising)
- no sign change → neither (a plateau, like x³ at 0)
A sign chart is just organized bookkeeping. Procedure:
- Find every x where f′ = 0 or f′ is undefined. Mark them on a number line.
- These points cut the line into intervals. Within each interval f′ cannot change sign — it would have to pass through zero or blow up to do so, and you've already found all those places.
- Pick any convenient test value inside each interval, plug into f′, record only the sign.
Example: f′(x) = (x−1)(x+2). Zeros at −2 and 1. Test x = −3: (−4)(−1) = + . Test x = 0: (−1)(2) = − . Test x = 2: (1)(4) = + . So f increases, decreases, increases: local max at −2, local min at 1.
Shortcut: you only need signs, so count negative factors rather than multiplying out.
Concavity and the second derivative
f″ > 0 says f′ is increasing — the slope itself is getting larger. A curve whose slope keeps rising (from steeply negative, through zero, to steeply positive) is by definition curving upward. The second derivative isn't a new kind of object; it's the first derivative rule applied one level up.
This is also exactly why the second derivative test works: at a critical point the slope is zero, and if the slope is increasing through zero, it went from negative to positive — falling then rising — which is a minimum.
Second derivative test
At a critical point c where f′(c) = 0:
- f″(c) < 0 → local maximum (concave down, so the flat spot is a peak)
- f″(c) > 0 → local minimum
- f″(c) = 0 → inconclusive; fall back to the first derivative test
x⁴, −x⁴, and x³ all have f′(0) = f″(0) = 0, and they have a minimum, a maximum, and neither, respectively. The test simply can't distinguish them.
What the test is actually doing: checking whether the best quadratic approximation to f near c opens up or down. When f″(c) = 0 the quadratic is flat and carries no information — you'd have to look at the cubic term. That's Taylor series (Unit 10) leaking backwards into November.
Curve analysis — the full procedure
- Domain, intercepts, asymptotes. Vertical where a denominator vanishes and the numerator doesn't; horizontal from the limit at ±∞; slant when the numerator's degree exceeds by exactly one.
- f′. Critical points: f′ = 0 or undefined, and in the domain.
- Sign chart for f′ → intervals of increase/decrease → local extrema.
- f″. Candidates: f″ = 0 or undefined.
- Sign chart for f″ → concavity → inflection points (verify the sign change).
Analyze f(x) = x³ − 3x² + 2.
| domain all reals; no asymptotes | polynomial |
| f′ = 3x² − 6x = 3x(x − 2) | factor immediately — sign charts need factors, not expansions |
| critical points x = 0, x = 2 | |
| test x=−1: 3(−1)(−3) = + · x=1: 3(1)(−1) = − · x=3: 3(3)(1) = + | sign chart for f′ |
| increasing (−∞,0), decreasing (0,2), increasing (2,∞) | |
| local max at x=0 (+ to −), value f(0) = 2 | first derivative test |
| local min at x=2 (− to +), value f(2) = 8−12+2 = −2 | |
| f″ = 6x − 6 = 6(x − 1) | |
| f″ < 0 for x<1, f″ > 0 for x>1 | sign actually changes |
| inflection point at (1, 0); concave down then up | f(1) = 1−3+2 = 0 |
Cross-check with the second derivative test: f″(0) = −6 < 0 → max ✓. f″(2) = 6 > 0 → min ✓. The two tests agreeing is a free error check.
Not "there's a max at x = 0." Write: "f′ changes from positive to negative at x = 0, therefore f has a local maximum there." The reason, not just the conclusion. Roughly a third of the free-response points in this unit are justification points, and students who can do all the calculus routinely lose them.
Same for concavity: "f″ > 0 on (1,∞), therefore f is concave up there."
Reading a graph of f′
An entire AP question type: you're shown the graph of the derivative and asked about the function. Everything shifts one level.
| On the graph of f′ | About f |
|---|---|
| above the axis | f is increasing |
| below the axis | f is decreasing |
| crosses + → − | local maximum of f |
| crosses − → + | local minimum of f |
| f′ is increasing (sloping up) | f is concave up |
| local extremum of f′ | inflection point of f |
| area between f′ and the axis, a to b | net change f(b) − f(a) (Unit 6) |
Confusing "f′ is increasing" with "f is increasing." They are different claims and both can be false while the other is true. f′ can be increasing while staying negative — meaning f is decreasing but decelerating. Say it out loud each time: the height of f′ tells me whether f rises; the slope of f′ tells me how f bends.
The graph of f′ is a parabola opening upward with zeros at x = 1 and x = 5, and f(0) = 3. Describe f.
| f′ = positive on (−∞,1), negative on (1,5), positive on (5,∞) | upward parabola sits above the axis outside its roots |
| so f increases, decreases, increases | height of f′ governs direction of f |
| local max at x = 1, local min at x = 5 | where f′ crosses, and in which direction |
| the parabola's vertex is at x = 3 | midpoint of the roots |
| f′ decreasing on (−∞,3), increasing on (3,∞) | slope of f′ |
| f is concave down on (−∞,3), concave up on (3,∞); inflection at x = 3 | note the inflection sits at f′'s minimum, not at its zeros |
The value f(0) = 3 shifts the whole curve vertically but changes none of the shape conclusions. That's the +C again, showing up as "the derivative determines f only up to a constant."
Optimization
The procedure
- Draw it and name variables.
- Write the objective — the quantity being maximized or minimized.
- Write the constraint — the relationship limiting your choices.
- Use the constraint to reduce the objective to one variable.
- State the domain. This decides whether endpoints are candidates.
- Differentiate; find critical points.
- Justify max or min (first or second derivative test, or the closed-interval method).
- Answer the question actually asked — sometimes dimensions, sometimes the optimal value.
The calculus in an optimization problem is trivial — differentiate a function and set it to zero, which she could do in week six. What's being tested is whether she can convert a paragraph of English into one function of one variable.
The structure is always the same: two quantities, one you're optimizing and one that's fixed. Find which is which. "Minimize material for a box holding 32 cubic units" — material is the objective, volume is the constraint. "Maximize area with 100 feet of fence" — area is the objective, perimeter is the constraint. Getting these backwards is the most common failure, and it happens before any calculus starts.
An open-top box with a square base must hold 32 cubic units. Minimize the material used.
| let base side = x, height = h | name what varies |
| objective: S = x² + 4xh | base plus four sides; no top, so no second x² |
| constraint: x²h = 32 | the fixed volume |
| h = 32/x² | solve the constraint for the easier variable |
| S = x² + 4x(32/x²) = x² + 128/x | substitute — now one variable |
| domain x > 0 | open interval, so no endpoints to check |
| S′ = 2x − 128/x² = 0 | rewrite 128/x as 128x−1 before differentiating |
| 2x³ = 128, x³ = 64, x = 4 | multiply through by x² |
| S″ = 2 + 256/x³ > 0 for x > 0 | justification: concave up everywhere, so this is the minimum |
| x = 4, h = 32/16 = 2 | answer the question: the dimensions |
Find the absolute max and min of f(x) = x³ − 3x on the closed interval [0, 3].
| f continuous on a closed bounded interval | EVT guarantees both exist |
| f′ = 3x² − 3 = 3(x−1)(x+1) | |
| critical points x = 1 and x = −1 | x = −1 is outside [0,3]; discard it |
| candidates: x = 0, x = 1, x = 3 | critical points inside, plus both endpoints |
| f(0) = 0 · f(1) = 1 − 3 = −2 · f(3) = 27 − 9 = 18 | evaluate f, not f′ |
| absolute max 18 at x = 3; absolute min −2 at x = 1 | the max is at an endpoint and is not a local max at all |
This is the closed interval method, and it's mechanical: list critical points inside plus both endpoints, evaluate f at each, compare. Forgetting the endpoints is the single most common lost point in the unit. EVT is what guarantees the comparison is exhaustive.
Next stop — the calculus of variations, which optimizes over a space of functions rather than numbers. The brachistochrone problem from Unit 9 is its founding question. The Euler–Lagrange equation is its central result.
Then — Lagrangian and Hamiltonian mechanics, which reformulate all of physics as "nature minimizes a certain integral." That reformulation is what makes quantum mechanics and field theory expressible at all. It's arguably the deepest idea reachable from a BC starting point.
If there is one thing worth importing from the Russian tradition, it's this: prove inequalities by monotonicity. The method is three steps, it needs nothing beyond this unit, and it converts a whole class of problems that look impossible into problems that are routine.
The recipe: to show A(x) ≥ B(x) on an interval, define f = A − B, show f′ has a definite sign, and check one endpoint value.
| Prove ex ≥ 1 + x for all real x. | |
| let f(x) = ex − 1 − x | difference of the two sides |
| f′(x) = ex − 1 | negative for x < 0, positive for x > 0 |
| so f decreases then increases; minimum at x = 0 | first derivative test |
| f(0) = 1 − 1 − 0 = 0, so f ≥ 0 everywhere. ∎ | the minimum value is zero, so the difference is never negative |
Why this is powerful: the inequality ex ≥ 1 + x is the tangent line at zero lying below a convex curve — and once you see that, you have a machine. Every convexity fact is an inequality; every inequality of that shape is a calculus problem.
Try these, in increasing difficulty:
- sin x < x for x > 0 — and notice this is the Unit 1 sector inequality, reproved without geometry
- ln(1+x) ≤ x for x > −1
- x − x³/6 ≤ sin x for x ≥ 0 — needs the method applied twice, differentiating down to a known inequality
- AM–GM for two terms: (a+b)/2 ≥ √(ab) — fix b, treat as a function of a
The third one is the first term of the Taylor series for sine, proved to be a lower bound eight months before Taylor series appears. That's the kind of connection this method keeps producing.
Convexity, stated properly. BC says "concave up" and draws a picture. The real definition doesn't mention derivatives at all:
In words: the chord lies above the curve. That's it — no differentiability required, which matters because plenty of important convex functions have corners (|x| is convex).
When f is twice differentiable, this is equivalent to f″ ≥ 0, which is the BC version. But the general definition is what generalizes: it's the foundation of convex optimization, which is the branch of applied mathematics that actually gets used — in economics, in control theory, in machine learning — precisely because convex problems are the ones that can be solved reliably at scale.
Jensen's inequality is the same statement for many points at once, and it implies AM–GM, Cauchy–Schwarz, and a good fraction of the inequalities in competition mathematics as special cases.
Unit 5 — Analysis of Functions
The theorem chain
- EVT: f continuous on a closed [a,b] ⟹ absolute max and min are attained
- Rolle: + f(a) = f(b) ⟹ some c with f′(c) = 0
- MVT: f continuous on [a,b], differentiable on (a,b) ⟹ some c with f′(c) = [f(b)−f(a)]/(b−a)
- Consequences: f′>0 ⟹ increasing · f′=0 everywhere ⟹ constant · same derivative ⟹ differ by a constant (this is the +C)
Extrema
- Critical point: f′ = 0 or undefined, and in the domain
- Fermat: interior extremum ⟹ critical point. Converse false (x³ at 0)
- 1st derivative test: + to − = max · − to + = min · no change = neither
- 2nd derivative test: f′(c)=0 and f″(c)<0 = max · f″(c)>0 = min · f″(c)=0 = inconclusive
Shape
- f″ > 0 ⟺ f′ increasing ⟺ concave up ⟺ tangents lie below the curve
- Inflection point: f″ changes sign. f″ = 0 alone is not enough.
Optimization
- objective + constraint → one variable → differentiate → critical points → justify → answer what was asked
- Closed interval method: evaluate f at all interior critical points and both endpoints; compare. EVT guarantees this is exhaustive.
- Open domain ⇒ no endpoints, but say so.
Procedure
- domain/asymptotes → f′ → sign chart → extrema → f″ → sign chart → concavity and inflections
- Factor derivatives before building a sign chart; you only ever need signs
Graph of f′ → f
- Height of f′ ⟹ whether f rises or falls
- Slope of f′ ⟹ how f bends
- Zeros of f′ with a sign change ⟹ extrema of f · extrema of f′ ⟹ inflections of f
AP language that scores
- "f′ changes from positive to negative at x = c, therefore f has a local maximum at c."
- "f is continuous on [a,b] and differentiable on (a,b), so by the Mean Value Theorem…"
- Always state hypotheses before invoking EVT, IVT, or MVT.
An optimum has to beat the endpoints. Fencing a rectangle with 20 m of perimeter gives A = x(10 − x), maximised at x = 5 with A = 25. The endpoints x = 0 and x = 10 both give 0, so 25 really is the maximum and not a minimum you have misidentified.
Two seconds of arithmetic, and it catches a flipped sign in the second-derivative test.
Practise Unit 5 → Fresh problems from this unit, with worked solutions.