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Maxima and minima

Of all rectangles with perimeter 40, the one with the largest area is the 10 by 10 square. The derivative finds it by looking for a slope of zero.

Peaks, valleys and critical points

A function has a local maximum at c if f(c) is at least as large as every nearby value, and a local minimum if it is at least as small. These are the peaks and valleys of the graph. Fermat's theorem, as OpenStax states it, says: if f has a local extremum at c and f is differentiable at c, then f′(c) = 0. The reason is in the picture: at a smooth peak the tangent is horizontal. So the candidates for local extrema are the critical points, the values c where f′(c) = 0 or f′(c) does not exist.

A critical point is only a candidate. f(x) = x³ has f′(0) = 0, but the graph rises on both sides of 0 and has no peak or valley there. To decide, use the first derivative test: if f′ changes from positive to negative at c, f has a local maximum there; from negative to positive, a local minimum; no change of sign, neither. For f(x) = x³ − 3x, f′(x) = 3x² − 3 = 0 gives x = −1 and x = 1. The derivative goes +, −, + across them, so f(−1) = 2 is a local maximum and f(1) = −2 is a local minimum.

f(x) = x³ − 3x. Horizontal tangents at the critical points x = ±1. On [0, 3] the lowest value is −2 at x = 1 and the highest is 18 at the endpoint x = 3.

Highest and lowest on a closed interval

The extreme value theorem guarantees that a continuous function on a closed interval [a, b] has an absolute maximum and an absolute minimum there. They occur either at critical points inside the interval or at the endpoints. That gives a short procedure, the closed interval method: find the critical points in (a, b), evaluate f at each of them and at a and b, then pick the largest and smallest values. For x³ − 3x on [0, 3]: f(0) = 0, f(1) = −2, f(3) = 18. The absolute minimum is −2 and the absolute maximum is 18, at an endpoint where f′ is not zero at all.

An optimization problem

Here is a made-up example in the style of OpenStax section 4.7. You have 40 meters of fence for a rectangular pen. If one side is x, the other is 20 − x, so the area is A(x) = x(20 − x) = 20x − x² for 0 ≤ x ≤ 20. A′(x) = 20 − 2x is zero at x = 10. The endpoints give A(0) = A(20) = 0, and A(10) = 100. So the best pen is a 10 by 10 square with 100 square meters. The steps are always the same: write the quantity as a function of one variable, find its interval, then use the closed interval method.

Corners count. |x| has its minimum at 0, where f′ does not exist. That is why critical points include places where the derivative fails to exist, not only places where it equals zero.
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f(x) = x³ − 3x with its critical points marked. Drag the point over the peak and the valley: the slope passes through 0 at each.
  1. Critical points of f(x) = x³ − 3x?
    • x = −1 and x = 1
    • x = 0 only
    • x = 3
    • x = ±3
  2. Local maximum value of x³ − 3x?
  3. Local minimum value of x³ − 3x?
  4. Absolute maximum of x³ − 3x on [0, 3]?
  5. f(x) = x² − 6x + 5. Minimum value?
  6. Same f on [0, 4]: absolute maximum?
  7. A rectangle has perimeter 40. Largest possible area?
  8. Perimeter 60 instead. Largest area?
  9. f(x) = x³ has f′(0) = 0. At x = 0, f has:
    • neither a max nor a min
    • a local max
    • a local min
    • an absolute max
  10. f′ changes from negative to positive at c. Then f(c) is:
    • a local minimum
    • a local maximum
    • neither
    • undefined
  11. Where can the absolute extremes of a continuous f on [a, b] occur?
    • At critical points or at the endpoints
    • Only where f′ = 0
    • Only at the endpoints
    • Only at x = 0
Key points
  • Critical point: f′(c) = 0 or f′(c) does not exist.
  • Fermat: a local extremum where f is differentiable has f′ = 0; the converse fails (x³).
  • First derivative test: + to − is a max, − to + is a min.
  • Closed interval method: compare f at critical points and endpoints.
Sources for this course (10)
  • OpenStax, Calculus Volume 1 : 2.1 A Preview of Calculus (tangent as the limit of secants; rock s(t) = −16t² + 64; stone h(t) = 15t − 4.9t² in the exercises: openstax.org/books/calculus-volume-1/pages/2-1-a-preview-of-calculus)
  • chapter 3 Derivatives: 3.1 Defining the Derivative (difference quotient, f′(a) = lim (f(a + h) − f(a))/h, tangent line: .../pages/3-1-defining-the-derivative)
  • 3.2 The Derivative as a Function (graphing f′, differentiability implies continuity, |x| has a corner at 0, higher derivatives: .../pages/3-2-the-derivative-as-a-function)
  • 3.3 Differentiation Rules (constant, power, sum, difference, constant multiple, product and quotient rules: .../pages/3-3-differentiation-rules)
  • 3.4 Derivatives as Rates of Change (position, velocity, speed, acceleration: .../pages/3-4-derivatives-as-rates-of-change)
  • 3.5 Derivatives of Trigonometric Functions (d/dx sin x = cos x using lim (sin h)/h = 1 and lim (cos h − 1)/h = 0: .../pages/3-5-derivatives-of-trigonometric-functions)
  • 3.6 The Chain Rule (.../pages/3-6-the-chain-rule)
  • chapter 4: 4.3 Maxima and Minima (critical points, Fermat's theorem, extreme value theorem, closed interval method: .../pages/4-3-maxima-and-minima)
  • 4.5 Derivatives and the Shape of a Graph (first derivative test: .../pages/4-5-derivatives-and-the-shape-of-a-graph)
  • 5.3 for Newton and Leibniz working in the late 1600s and early 1700s (.../pages/5-3-the-fundamental-theorem-of-calculus).

Checked: 2026-10-17