Every rule in Weeks 4-5 computed a derivative. This week asks what a derivative is for: reading it tells you where a function is largest, smallest, rising, falling, curving up, or curving down — without ever plotting a single point by hand.
6.1 Absolute and Local Extreme Values
f has an absolute maximum at c if f(c) ≥ f(x) for every x in the domain; an absolute minimum is defined the same way with ≤. A local maximum requires f(c) ≥ f(x) only for x near c — every absolute extremum that occurs at an interior point is automatically a local one, but not conversely.
6.2 The Extreme Value Theorem and Critical Points
If f is continuous on a closed interval [a,b], it attains both an absolute maximum and an absolute minimum somewhere in [a,b] — the Extreme Value Theorem. Continuity and closedness are both essential: an open interval or a discontinuity can let a function approach a supremum it never actually reaches.
Local extrema can only occur at critical points: points c in the domain where f′(c) = 0 or f′(c) does not exist. This is necessary but not sufficient — not every critical point is an extremum (e.g. f(x)=x3 has f′(0)=0 but no extremum at 0).
6.3 The Closed Interval Method
To find the absolute extrema of a continuous f on [a,b]:
- Find all critical points of f in (a,b).
- Evaluate f at each critical point and at both endpoints a, b.
- The largest value found is the absolute maximum; the smallest is the absolute minimum.
Example 6.1 — The Closed Interval Method
Find the absolute extrema of f(x) = x3 − 3x on [−2, 3].
Step 1 — Find critical points
f′(x) = 3x2−3 = 0 ⇒ x = ±1, both in (−2,3).
Step 2 — Evaluate f at critical points and endpoints
f(−2)=−2; f(−1)=2; f(1)=−2; f(3)=18
Step 3 — Compare
Absolute maximum 18 at x=3; absolute minimum −2, attained at both x=−2 and x=1.
6.4 Rolle's Theorem and the Mean Value Theorem
Rolle's Theorem. If f is continuous on [a,b], differentiable on (a,b), and f(a)=f(b), then f′(c)=0 for some c in (a,b).
The Mean Value Theorem (MVT) generalises Rolle's Theorem by tilting the picture: if f is continuous on [a,b] and differentiable on (a,b), then
Geometrically: at some interior point, the tangent line is parallel to the secant line joining the endpoints (Week 2 §2.2). Rolle's Theorem is the special case where that secant is horizontal.
Example 6.2 — Applying the Mean Value Theorem
Verify the MVT for f(x) = x2 on [0, 4] by finding c explicitly.
Step 1 — Compute the average rate of change (right side of Eq. 6.1)
(16−0)/(4−0) = 4.
Step 2 — Set f'(c) equal to it and solve
f′(x) = 2x, so 2c = 4.
c = 2, which does lie in (0,4) as required.
6.5 Consequences of the Mean Value Theorem
Two facts, both proved directly from the MVT, are used constantly:
- If f′(x) = 0 for every x in an interval, f is constant on that interval.
- If f′(x) = g′(x) for every x in an interval, then f(x) = g(x) + C for some constant C.
The second fact is the reason antiderivatives (Week 9) always come in families differing only by a constant — a fact you will rely on throughout integration.
6.6 Monotonic Functions and the First Derivative Test
| Sign of f′ on an interval | Behaviour of f |
|---|---|
| f′(x) > 0 | f is increasing |
| f′(x) < 0 | f is decreasing |
This is a direct consequence of the MVT: if f′ > 0 throughout an interval, any two points there have a secant slope equal to some interior f′(c) > 0, so the function must rise. The First Derivative Test reads local extrema directly from a sign change of f′ at a critical point c:
- f′ changes from + to − at c: local maximum at c.
- f′ changes from − to + at c: local minimum at c.
- f′ does not change sign at c: no local extremum there.
Example 6.3 — The First Derivative Test
Find and classify the local extrema of f(x) = x3 − 3x2.
Step 1 — Critical points
f′(x) = 3x2−6x = 3x(x−2) = 0 ⇒ x = 0, 2.
Step 2 — Test the sign of f' on each interval
On (−∞,0): pick x=−1, f′(−1)=9>0.
On (0,2): pick x=1, f′(1)=−3<0.
On (2,∞): pick x=3, f′(3)=9>0.
Step 3 — Classify using the sign changes
+ to − at x=0: local maximum, f(0)=0. − to + at x=2: local minimum, f(2)=−4.
6.7 Concavity and the Second Derivative
The second derivative f″(x) is the derivative of f′(x). Its sign describes concavity:
| Sign of f″ on an interval | Shape of the graph |
|---|---|
| f″(x) > 0 | Concave up (like a cup; f′ is increasing) |
| f″(x) < 0 | Concave down (like a cap; f′ is decreasing) |
6.8 Points of Inflection
An inflection point is a point on the graph where concavity changes sign. At an inflection point, f″(x)=0 or f″(x) fails to exist — but as with critical points and extrema, this condition is necessary, not sufficient: check that concavity actually changes sign on either side.
6.9 The Second Derivative Test
At a critical point c where f′(c)=0:
- If f″(c) > 0, f has a local minimum at c (concave up at the bottom of a dip).
- If f″(c) < 0, f has a local maximum at c (concave down at the top of a hump).
- If f″(c) = 0, the test is inconclusive — fall back on the First Derivative Test.
Example 6.4 — The Second Derivative Test
Classify the critical points of f(x) = x3 − 3x2 using f″.
Step 1 — Find f″
f′(x) = 3x2−6x; f″(x) = 6x−6.
Step 2 — Evaluate at each critical point (x=0, x=2 from Example 6.3)
f″(0) = −6 < 0; f″(2) = 6 > 0.
Step 3 — Conclude
Local maximum at x=0 (matches Example 6.3); local minimum at x=2 (matches Example 6.3) — the two tests agree, as they always do when both apply.
6.10 A Curve-Sketching Strategy
- Find the domain and any symmetry (Week 1 §1.4) or asymptotes (Week 3 §3.7–3.8).
- Find f′, its critical points, and where f is increasing/decreasing (§6.6).
- Classify each critical point with the First or Second Derivative Test (§6.6, §6.9).
- Find f″, its zeros, and intervals of concavity; identify inflection points (§6.7–6.8).
- Plot a few key points (intercepts, extrema, inflection points) and sketch, guided by all of the above.
6.11 Applied Optimization
Optimization problems ask for the largest or smallest value of some quantity subject to a constraint. The strategy:
- Draw a picture and name every relevant variable.
- Write down the quantity to optimize as a function of one variable, using the constraint equation to eliminate all others.
- Determine the domain of that function from the physical context.
- Use the Closed Interval Method (§6.3) or the First/Second Derivative Test to find the extreme value.
Example 6.5 — Maximizing enclosed area
A farmer has 200 m of fencing to enclose a rectangular field against an existing straight wall (so only three sides need fencing). Find the dimensions that maximize the enclosed area.
Step 1 — Name variables and write the constraint
Let x = length of each of the two sides perpendicular to the wall, y = length of the side parallel to the wall. Fencing used: 2x + y = 200.
Step 2 — Write the quantity to optimize as a function of one variable
y = 200 − 2x, so Area A(x) = xy = x(200−2x) = 200x − 2x2, with domain 0 < x < 100.
Step 3 — Find the critical point
A′(x) = 200 − 4x = 0 ⇒ x = 50.
Step 4 — Confirm it is a maximum and state the dimensions
A″(x) = −4 < 0 for every x, so by the Second Derivative Test x=50 gives a maximum.
x = 50 m, y = 200−100 = 100 m, giving maximum area A = 50(100) = 5000 m².
6.12 Nomenclature
| Symbol | Meaning |
|---|---|
| critical point | A point c where f'(c)=0 or f'(c) does not exist |
| MVT | Mean Value Theorem |
| f″(x) | The second derivative of f, i.e. (f')' |
| inflection point | A point where concavity changes sign |
| concave up / down | f″>0 (cup-shaped) / f″<0 (cap-shaped) |
6.13 Problems
- Closed Interval Method. Find the absolute extrema of f(x) = x3 − 12x on [−3, 3].
- Critical points. Find all critical points of f(x) = x4 − 4x2.
- Mean Value Theorem. Find the value of c guaranteed by the MVT for f(x) = x3 on [0, 2].
- Consequence of the MVT. Two functions satisfy f′(x) = g′(x) for all x, and f(0)=3, g(0)=1. Find f(x) − g(x) for every x.
- First Derivative Test. Find and classify all local extrema of f(x) = x3 − 3x + 1.
- Concavity. Find the intervals of concavity and any inflection points of f(x) = x4 − 6x2.
- Second Derivative Test. Use the Second Derivative Test to classify the critical points of f(x) = x3 − 3x + 1 (from Problem 5), and confirm the two methods agree.
- Curve sketching. Sketch f(x) = x4 − 4x3, identifying all extrema and inflection points.
- Optimization, area. A rectangular box with a square base and open top must have volume 32 m³. Find the dimensions that minimize the surface area.
- Optimization, distance. Find the point on the line y = 2x + 1 closest to the point (3, 0).
- Optimization, revenue. A company sells q units at a price of p(q) = 50 − 0.5q per unit. Find the value of q that maximizes revenue R(q) = qp(q).
6.14 References
- Thomas, G. B., Weir, M. D. and Hass, J. Thomas' Calculus, SI Units, 15th ed. Pearson, 2024. — §4.1 (extreme values of functions), §4.2 (the Mean Value Theorem), §4.3 (monotonic functions and the first derivative test), §4.4 (concavity and curve sketching), §4.5 (applied optimization).
- Pearson MyLab Math — the Week 6 homework set is assigned directly from the sections above; see the Calendar for the due date.