Difficulty: Intermediate | Prerequisites: Basic derivative rules (power, chain, product, quotient), critical numbers from Ch. 4.1
Big Picture
This section answers the question: on a closed interval, where does a function reach its highest and lowest values? It builds directly on the idea of critical numbers from earlier in the chapter and adds the step of checking endpoints. In applied problems (pricing, cost, area), this is how you find the best or worst outcome within a constrained range. You need to be comfortable taking derivatives and solving f'(x) = 0 before diving in.
To find the absolute max and min of a continuous function on a closed interval [a, b], take the derivative, find critical numbers inside the interval, evaluate f at those points and at both endpoints, then compare. The largest value is the absolute maximum; the smallest is the absolute minimum. The Extreme Value Theorem guarantees both exist on a closed interval.
Absolute maximum (global maximum)
The largest value a function attains on a given interval. On a closed interval, it must occur at a critical number or at an endpoint.
Think of it as: the highest point on the curve between the two fences.
Absolute minimum (global minimum)
The smallest value a function attains on a given interval. Like the absolute max, it occurs at a critical number or an endpoint.
Think of it as: the lowest point on the curve between the two fences.
Critical number (critical point)
A value x = c in the domain of f where f'(c) = 0 or f'(c) does not exist. These are the only interior candidates for extrema.
In simple terms, this means: wherever the slope is zero or undefined, something interesting might be happening.
Closed interval [a, b]
An interval that includes both of its endpoints. Written with square brackets. The Extreme Value Theorem requires a closed interval.
Think of it as: the fence posts are included in the yard.
Extreme Value Theorem (EVT)
If f is continuous on a closed interval [a, b], then f has both an absolute maximum and an absolute minimum on that interval.
In simple terms, this means: a continuous curve on a closed interval is guaranteed to have a highest and lowest point. No exceptions.
The Closed Interval Method
This is the workhorse procedure for finding absolute extrema on [a, b]. Every problem in this section uses it.
Step 1: Find f'(x).
Step 2: Set f'(x) = 0 and solve. Also note where f'(x) is undefined (but f(x) is defined). These give the critical numbers.
Step 3: Discard any critical numbers that fall outside [a, b].
Step 4: Evaluate f at each remaining critical number and at both endpoints a and b.
Step 5: The largest value from Step 4 is the absolute maximum. The smallest is the absolute minimum.
Worked Example: f(x) = 5 ln(x) - 4x on [1, 3]
f'(x) = 5/x - 4. Set equal to zero: 5/x = 4, so x = 5/4. This is in [1, 3], so it stays.
Evaluate: f(1) = 5 ln(1) - 4(1) = -4. f(5/4) = 5 ln(5/4) - 5 = approx. -3.884. f(3) = 5 ln(3) - 12 = approx. -6.507.
Absolute maximum: -3.884 at x = 5/4. Absolute minimum: -6.507 at x = 3.
Worked Example: f(x) = x^4 - 6x^2 on [1, 3]
f'(x) = 4x^3 - 12x = 4x(x^2 - 3). Critical numbers: x = 0, x = sqrt(3), x = -sqrt(3). Only x = sqrt(3) is in [1, 3].
Evaluate: f(1) = 1 - 6 = -5. f(sqrt(3)) = 9 - 18 = -9. f(3) = 81 - 54 = 27.
Absolute maximum: 27 at x = 3. Absolute minimum: -9 at x = sqrt(3).
Worked Example: f(x) = (x + 4)(x - 2)^2 on [-1, 3]
Expand or use the product rule to find f'(x), then set it to zero. Critical number x = 2 is in [-1, 3] (and x = 0 if using the expanded form).
Evaluate at critical numbers and endpoints. f(-1) = (3)(9) = 27. f(2) = (6)(0) = 0. f(3) = (7)(1) = 7.
Absolute maximum: 27 at x = -1. Absolute minimum: 0 at x = 2.
Functions involving e^x and ln(x)
The same method applies. For f(x) = 5xe^x on [0, 2]:
f'(x) = 5e^x + 5xe^x = 5e^x(1 + x). Setting equal to zero: 1 + x = 0, so x = -1. But -1 is not in [0, 2], so there are no interior critical numbers.
Evaluate endpoints: f(0) = 0 and f(2) = 10e^2.
Absolute minimum: 0 at x = 0. Absolute maximum: 10e^2 at x = 2.
Rational functions on closed intervals
For f(x) = 2x/(x + 3) on [1, 4], use the quotient rule. f'(x) = 6/(x + 3)^2, which is always positive on the interval, so no critical numbers from f'(x) = 0. Just compare endpoints: f(1) = 1/2 and f(4) = 8/7.
Absolute minimum: 1/2 at x = 1. Absolute maximum: 8/7 at x = 4.
Power rule: d/dx [x^n] = n x^(n-1)
Product rule: d/dx [u v] = u' v + u v'
Quotient rule: d/dx [u/v] = (u' v - u v') / v^2
Chain rule: d/dx [f(g(x))] = f'(g(x)) g'(x)
d/dx [ln(x)] = 1/x
d/dx [e^x] = e^x
The closed interval method itself is not a formula but a procedure. The formulas above are the toolkit for Step 1 (finding f'(x)).
Students often forget to check the endpoints. A critical number gives you candidates, but the absolute extrema on a closed interval can occur at x = a or x = b, not just where f'(x) = 0.
Students sometimes include critical numbers that fall outside the interval. If your interval is [1, 3] and you find x = -1 as a critical number, throw it out.
Confusing "absolute" with "local." An absolute maximum is the single highest value on the entire interval. A local maximum is just a hilltop in the neighbourhood. On a closed interval, the absolute max might be at an endpoint, which would not count as a local max.
Setting f(x) = 0 instead of f'(x) = 0. You need the derivative equal to zero, not the function itself.
The closed interval method appears on nearly every MAC 2233 exam. Expect at least one or two problems requiring it.
The exam frequently includes functions with ln(x), e^x, and rational expressions, so be comfortable differentiating those.
Reading absolute extrema from a graph (as in Problem 13) is a common exam question. Remember: on a closed interval, both endpoints are included; on an open interval, the endpoint value may not count as an extremum.
When the problem says "find the absolute extrema," always state both the value and where it occurs (the x-value).
True or False: A continuous function on a closed interval is guaranteed to have both an absolute maximum and an absolute minimum. (True)
Fill in the blank: To find critical numbers, set ______ equal to zero. (f'(x))
True or False: If f'(x) = 0 at x = 5 but the interval is [0, 3], you should still evaluate f(5). (False, discard it)
True or False: The absolute maximum of a function on [a, b] can occur at an endpoint. (True)
Fill in the blank: After finding all critical numbers inside [a, b], you must also evaluate f at ______ and ______. (a, b)
Q: Find the absolute extrema of f(x) = x^3 - 27x + 1 on [-4, 0].
A: f'(x) = 3x^2 - 27 = 3(x^2 - 9) = 3(x - 3)(x + 3). Critical numbers: x = 3 and x = -3. Only x = -3 is in [-4, 0]. Evaluate: f(-4) = -64 + 108 + 1 = 45, f(-3) = -27 + 81 + 1 = 55, f(0) = 1. Absolute maximum is 55 at x = -3. Absolute minimum is 1 at x = 0.
Q: Find the absolute extrema of f(x) = 2x/(x + 3) on [1, 4].
A: f'(x) = 6/(x + 3)^2. This is never zero (numerator is a constant), so no critical numbers. Compare endpoints: f(1) = 2/4 = 1/2, f(4) = 8/7. Absolute minimum is 1/2 at x = 1. Absolute maximum is 8/7 at x = 4.
Q: Find the absolute extrema of f(x) = 5xe^x on [0, 2].
A: f'(x) = 5e^x(1 + x). Setting 1 + x = 0 gives x = -1, which is outside [0, 2]. Compare endpoints: f(0) = 0, f(2) = 10e^2. Absolute minimum is 0 at x = 0. Absolute maximum is 10e^2 at x = 2.
Q: On the graph of a function, the interval is [0, 7]. You read the following y-values: at x = 0, y = 0; at x = 4, y = 5 (a peak); at x = 7, y = 1. What are the absolute extrema?
A: Absolute minimum is 0 at x = 0. Absolute maximum is 5 at x = 4.
This connects to Ch. 4.6 (second derivative test) because the second derivative test is an alternative way to classify critical points, especially on open intervals where the closed interval method does not apply. It also connects to applied optimization (Ch. 4.6 word problems), where you first build the function and constraint, then use these same techniques to find the optimal value. Later, in Ch. 5.5, you will use definite integrals to compute average values of functions over intervals, which is a different kind of "best summary" of a function's behaviour on [a, b].
absolute extrema, global extrema, closed interval method, Extreme Value Theorem, EVT, critical numbers, critical points, endpoint evaluation, absolute max, absolute min, calculus for business, MAC 2233, FIU, Ch 4.5, continuous function extrema