Maximum and Minimum Values
Absolute Maximum / Minimum
Definition: A function f has an absolute maximum (also called global
maximum) at c if f (c) ≥ f (x) for all x in its domain, D. The value f (c) is
called the maximum value of f. A function f has an absolute minimum (or
global minimum) at c if f (c) ≤ f (x) for all x in its domain. Such a value f (c)
is called the minimum value of f. The maximum and minimum values of f
are called the extreme values of f.
Local Maximum / Minimum
Definition: A function f has a local maximum (or relative maximum) at c if
f (c) ≥ f (x) when x near c. That is, f (c) ≥ f (x) for all x on some open
interval containing c. Similarly, f has a local minimum (or relative
minimum) at c if f (c) ≤ f (x) when x near c.
The Extreme Value Theorem (the existence of absolute maximum /
minimum): If f is continuous on a closed interval [a, b] , then there exist (at
least) a point c where f attains its maximum value, f (c), on the interval, and
(at least) a point d where f attains its minimum value, f (d), on the interval.
Comment: This means that if both of the following conditions: (1.)
the interval is closed, and (2.) f is continuous on it, are met, then f is
guaranteed to have (at least) one absolute maximum and one absolute
minimum points on the interval. If either condition fails, then the
existence of max / min points is not guaranteed.
Fermat’s Theorem (test for local extreme values): If f has a local
maximum or minimum at c, and if f ′(c) exists, then f ′(c) = 0.
Note 1: Therefore, it follows that a local extreme point can only occur at
places where either f ′(x) = 0 or f ′(x) is undefined (i.e., either at a point
where the tangent line is horizontal, or at a nondifferentiable point).
Examples: f (x) = (x − 2)2, at x = 2; g(x) =│x│, at x = 0.
Note 2: The converse is not always true: the fact that f ′(c) = 0, or that f ′(c)
does not exist, does NOT guarantee that c is a local extreme point of f.
Example: f (x) = x3, at x = 0.
Definition: A critical point or critical number of a function f is a point
x = c in the domain of f such that either f ′(c) = 0 or f ′(c) does not exist.
Comment: The critical points are all the candidate points for local
maximum / minimum of f . That is, every local extreme point is a
critical point, but not every critical point is a local extreme point.
Naturally, the max / min points of f have to be in the domain of f , i.e.
they are points on the graph of f. Therefore, for example, if f is
undefined at an infinite discontinuity then the point of discontinuity is
not a critical point even though f ′ does not exist there.
How to find the absolute maximum and minimum values of a continuous
function f on a closed interval:
1. Find all critical points of f in the given interval.
2. Evaluate f at the critical point(s) found in step 1, as well as at the
two endpoints of the interval.
3. The point(s) of the largest value of f is the absolute maximum(s),
the point(s) of the smallest value is the absolute minimum(s).
Ex. Find the absolute maximum and minimum points of f (x) = 4 − x2 on
each of the intervals (i.) [−3, 1], and (ii.) [2, 5].
f ′(x) = −2x
f ′ = 0 at x = 0 which is the only critical point because f is a
polynomial, therefore, it has no nondifferentiable points.
(i.) Evaluate f at the critical point 0 and the endpoints −3 and 1:
f (−3) = −5, f (0) = 4,
f (1) = 3
Therefore, the absolute maximum point is (0, 4), and the absolute
minimum point is (−3, −5).
(ii.) The critical point x = 0 is not in this interval, therefore, just evaluate f
at the endpoints 2 and 5:
f (2) = 0, f (5) = −21
Therefore, the absolute maximum point is (2, 0), and the absolute
minimum point is (5, −21).
Ex. Find the absolute maximum and minimum values of f (x) = x3 − 27x + 8
on the interval [0, 4].
Ans. abs max value of 8 at x = 0, abs min value of −46 at x = 3
Ex. Find the absolute maximum and minimum values of g(t) = t 3/5 on the
interval [−32, 1].
Ans. abs max value of 1 at t = 1, abs min value of −8 at t = −32