Source: MTH 132 Chapter 3 Lecture Notes, Michigan State University
Difficulty: Intermediate | Prerequisites: Limits, continuity, basic derivative rules (Chapter 2).
This topic is the gateway to Chapter 3, where calculus shifts from computing derivatives to using them. You are learning to find the highest and lowest values a function reaches, either on its entire domain or near a particular point. This matters because optimisation problems (finding the best, cheapest, fastest, largest outcome) all reduce to finding maximum or minimum values. You should be comfortable with continuity, the definition of the derivative, and basic differentiation rules before diving in.
Every continuous function on a closed interval [a, b] is guaranteed to hit an absolute maximum and an absolute minimum somewhere on that interval. To find them, you compute the derivative, solve for its zeros and undefined points (the critical numbers), evaluate the function at each critical number inside the interval and at both endpoints, then compare.
Absolute (global) maximum
f(c) is the absolute maximum of f on D if f(c) >= f(x) for every x in D. In simple terms, it is the single highest output value the function reaches on the entire domain.
Absolute (global) minimum
f(c) is the absolute minimum of f on D if f(c) <= f(x) for every x in D. Think of it as the lowest point on the whole curve over the domain.
Local maximum
f(c) is a local maximum if f(c) >= f(x) for all x near c (in some open interval containing c). In simple terms, it is a peak in the neighbourhood, even if it is not the highest point overall.
Local minimum
f(c) is a local minimum if f(c) <= f(x) for all x near c. Think of it as a valley relative to nearby points.
Extreme values (extrema)
A collective term for maximums and minimums, whether local or absolute.
Critical number
A number c in the domain of f where either f'(c) = 0 or f'(c) does not exist (DNE). These are the only interior points where an absolute extremum can occur. Think of it as any x-value where the slope is flat or the derivative breaks down.
Closed interval
An interval [a, b] that includes both of its endpoints. This is the setting where the Extreme Value Theorem applies.
Extreme Value Theorem (EVT)
If f is continuous on a closed interval [a, b], then f attains an absolute maximum value and an absolute minimum value somewhere in [a, b]. In plain terms, a continuous curve on a closed interval must have a highest and a lowest point.
An absolute max/min is the largest/smallest value of f over the entire domain D.
A local max/min is the largest/smallest value of f in some open interval around the point.
Every absolute extremum is also a local extremum (unless it occurs at an endpoint), but a local extremum need not be the absolute one.
Maximums and minimums must be finite real numbers. If f(x) tends to infinity, it does not count as a maximum.
The textbook uses "near c" to mean "in some open interval containing c." Because endpoints do not sit inside an open interval within the domain, endpoints technically cannot be local extrema under this definition.
Two conditions can prevent a function from having an absolute max or min on an interval:
The function is not continuous on the interval. Example: f(x) = 1/x on [-2, 2]. The function blows up near x = 0, so it is not continuous there. As x approaches 0 from the right the values increase without bound, and no finite minimum exists.
The interval is not closed. Example: f(x) = x on (0, 2]. The left endpoint is excluded. You can pick x = 0.001, then x = 0.0001, then x = 0.000000001, always finding a smaller value. There is no smallest value the function takes, so no absolute minimum exists.
The Extreme Value Theorem tells you exactly when you are guaranteed both an absolute max and an absolute min: the function must be continuous on a closed interval [a, b]. If either condition fails, the theorem does not apply and extrema may or may not exist.
Absolute extrema on a closed interval can only occur at two kinds of places: critical numbers inside the interval, or the endpoints. A critical number is any value c in the domain of f where f'(c) = 0 or f'(c) does not exist.
How to find critical numbers:
Compute f'(x).
Set f'(x) = 0 and solve for x. These give critical numbers where the tangent line is horizontal.
Find any x-values where f'(x) is undefined but f(x) is defined. These give critical numbers where the derivative breaks (cusps, corners, vertical tangents).
Example: f(x) = 2x^3 - 3x^2 - 12x + 5
f'(x) = 6x^2 - 6x - 12
Set 0 = 6(x^2 - x - 2) = 6(x - 2)(x + 1)
Critical numbers: x = 2 and x = -1
To find the absolute max and min of a continuous function f on [a, b]:
Find all critical numbers of f that lie in the open interval (a, b).
Evaluate f at each critical number and at both endpoints a and b.
The largest value is the absolute maximum; the smallest is the absolute minimum.
Example 1: f(x) = x^3 - 12x + 1 on [1, 4]
f'(x) = 3x^2 - 12, so 0 = 3(x^2 - 4), giving x = 2 and x = -2. Only x = 2 lies in (1, 4).
f(1) = 1 - 12 + 1 = -10
f(2) = 8 - 24 + 1 = -15
f(4) = 64 - 48 + 1 = 17
Absolute max = 17 at x = 4. Absolute min = -15 at x = 2.
Example 2: g(t) = t - 3t^(2/3) on [-1, 27]
g'(t) = 1 - 2t^(-1/3) = 1 - 2/(t^(1/3))
g'(t) is undefined at t = 0 (critical number). Setting g'(t) = 0 gives t^(1/3) = 2, so t = 8 (critical number).
g(0) = 0, g(8) = 8 - 3(4) = -4, g(-1) = -1 - 3(1) = -4, g(27) = 27 - 3(9) = 0
Absolute max = 0 at t = 0 and t = 27. Absolute min = -4 at t = 8 and t = -1.
Example 3: g(s) = 1/(s - 5) on [-5, -4]
g'(s) = -1/(s - 5)^2, which is never zero and never undefined on [-5, -4] (since s = 5 is outside the interval). No critical numbers in the interval.
g(-5) = 1/(-10) = -1/10. g(-4) = 1/(-9) = -1/9.
Absolute max = -1/10 at s = -5. Absolute min = -1/9 at s = -4.
Example 4: f(x) = (x - 1)(x - 5)^3 + 3 on [0, 6]
f'(x) = (x - 5)^3 + 3(x - 1)(x - 5)^2 = (x - 5)^2[(x - 5) + 3(x - 1)] = (x - 5)^2(4x - 8) = 4(x - 5)^2(x - 2)
Critical numbers: x = 5 and x = 2.
f(0) = (-1)(-5)^3 + 3 = 125 + 3 = 128
f(2) = (1)(-3)^3 + 3 = -27 + 3 = -24
f(5) = (4)(0)^3 + 3 = 3
f(6) = (5)(1)^3 + 3 = 8
Absolute max = 128 at x = 0. Absolute min = -24 at x = 2.
Closed Interval Method (for continuous f on [a, b]):
Compute f'(x).
Find all c in (a, b) where f'(c) = 0 or f'(c) DNE. These are the critical numbers.
Evaluate f(c) at each critical number and f(a), f(b) at the endpoints.
The largest output = absolute max. The smallest output = absolute min.
Critical number condition:
f'(c) = 0 or f'(c) does not exist, where c is in the domain of f.
Extreme Value Theorem (EVT):
f continuous on [a, b] implies f attains both an absolute maximum and an absolute minimum on [a, b].
Optimisation is everywhere: engineers design containers to minimise material cost, businesses set prices to maximise revenue, and physicists find equilibrium points by locating where energy is minimised. The closed interval method is the workhorse technique behind all of these problems.
Students often assume that every function has an absolute max and min. It does not. The EVT requires both continuity and a closed interval. Drop either condition and the guarantee vanishes.
Students sometimes forget to check the endpoints. The absolute max or min frequently occurs at an endpoint, not at a critical number.
A critical number is where f'(c) = 0 or f'(c) does not exist. Students sometimes only look for zeros of the derivative and miss points where the derivative is undefined.
Students sometimes confuse "critical number" with "extreme value." A critical number is a candidate. It does not guarantee a max or min occurs there.
The Extreme Value Theorem statement (hypotheses and conclusion) is a common exam question.
Expect problems where you must explain why the EVT does or does not apply (check continuity and whether the interval is closed).
The closed interval method is tested repeatedly: find critical numbers, evaluate at critical numbers and endpoints, compare.
Watch for functions with fractional exponents (e.g., t^(2/3)) where the derivative is undefined at t = 0. These give critical numbers that students miss.
When a problem gives a graph instead of a formula, you still apply the same logic: extrema occur at critical points (flat tangent or undefined derivative) or at endpoints.
True or False: If f is continuous on (0, 5], then f must have an absolute minimum on that interval.
Answer: False. The interval is not closed (0 is excluded), so the EVT does not apply.
Fill in the blank: A critical number of f is a value c in the domain of f where f'(c) = ____ or f'(c) ____.
Answer: 0; does not exist.
True or False: If f'(c) = 0, then f must have a local max or local min at x = c.
Answer: False. f'(c) = 0 makes c a critical number, but f could have neither a max nor a min there (e.g., an inflection point).
Fill in the blank: The Extreme Value Theorem requires f to be ____ on a ____ interval [a, b].
Answer: continuous; closed.
True or False: Endpoints of a closed interval should be evaluated when using the closed interval method.
Answer: True.
Q: Find the critical numbers of f(t) = 7t^(2/3) - 9t^(5/3).
A: f'(t) = (14/3)t^(-1/3) - 15t^(2/3) = (1/(3t^(1/3)))[14 - 15t]. f'(t) is undefined at t = 0 and equals zero when 14 - 15t = 0, i.e. t = 14/15. Critical numbers: t = 0 and t = 14/15.
Q: Find the absolute maximum and minimum of f(x) = x^3 - 12x + 1 on [1, 4].
A: f'(x) = 3x^2 - 12 = 3(x - 2)(x + 2). Critical numbers in (1, 4): x = 2. f(1) = -10, f(2) = -15, f(4) = 17. Absolute max = 17 at x = 4. Absolute min = -15 at x = 2.
Q: Explain why f(x) = 1/x has no absolute minimum on [-2, 2].
A: f is not continuous on [-2, 2] because it is undefined at x = 0. The EVT does not apply. As x approaches 0 from the right, f(x) increases without bound, and as x approaches 0 from the left, f(x) decreases without bound. No finite minimum exists.
Q: Does f(x) = x on (0, 2] have an absolute minimum? Why or why not?
A: No. The interval is not closed on the left. For any value you pick near 0, there is always a smaller positive number closer to 0. The function never attains a smallest value.
Q: Consider f(x) = (x - 1)(x - 5)^3 + 3 on [0, 6]. Find the absolute max and min.
A: f'(x) = 4(x - 5)^2(x - 2). Critical numbers: x = 5, x = 2. f(0) = 128, f(2) = -24, f(5) = 3, f(6) = 8. Absolute max = 128 at x = 0. Absolute min = -24 at x = 2.
This connects directly to optimisation problems (Chapter 3, Section 7), where you set up a function from a word problem and then use the closed interval method to find its maximum or minimum.
The concept of critical numbers reappears in the First Derivative Test and the Second Derivative Test, which classify critical points as local maxima, local minima, or neither.
The Extreme Value Theorem relies on continuity (Chapter 2). If you are shaky on where functions are continuous, revisit that material first.
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