Maximum and Minimum Values
Concepts: Suppose that f(x) is continuous on an interval Iand cis a point inside I.
(i) If f(x)≤f(c) (or f(x)≥f(c), respectively) for all xin I, then f(c) is an absolute maximum
(or an absolute minimum, respectively) of f(x) on I. An absolute maximum or an absolute
minimum of fis an absolute extremum of f.
(ii) If f(x)≤f(c) (or f(x)≥f(c), respectively) for all xin some open interval containing c, then
f(c) is a local maximum (or a local minimum, respectively) of f(x). A local maximum or a
local minimum of fis a local extremum of f.
(iii) A number cin the domain of fis a critical point of fif either f0(c) = 0 or f0(c) does not
exist.
(iv) (How to find extrema of a function?) Suppose that fis continuous on [a, b]. Then the
absolute extrema of fon [a, b] must occur at either a critical point inside (a, b), or at an end point
of the interval (x=aor x=b).
Example 1 Find all critical points of f(x) = x3−3x2+3x, and determine whether each represents
a local maximum, local minimum, or neither. (Page 267, #9)
Solution: Since f(x) is a polynomial, f(x) is differentiable (and so continuous) in its domain
(−∞,∞).
Compute
f0(x) = 3x2−6x+ 3 = 3(x2−2x+ 1) = 3(x−1)2.
Thus the only critical point of f(x) is x= 1, which divides the domain of f(x) into two intervals:
(−∞,1) and (1,∞).
Since f0(x)>0, when x6= 1, we conclude that f0(x)>0 in both (−∞,1) and (1,∞). Thus
y=f(x) is increasing in both intervals. Therefore, f(1) is neither a local maximum value nor a
local minimum value of f(x) in its domain.
Example 2 Find all critical points of f(x) = x3
4−4x1
4, and determine whether each represents
a local maximum, local minimum, or neither. (Page 267, #13)
Solution: Since f(x) is a linear combination of power functions, f(x) is continuous on its domain
[0,∞) and differentiable in (0,∞).
Compute
f0(x) = 3
4x−1
4−x−3
4=3x2
4−4
4x3
4
.
As x= 0 is in the domain of f(x) and f0(0) does not exist, thus x= 0 is a critical point. Set
f0(x) = 0 to get x2
4=4
3, or x=16
9. Hence the critical points of f(x) are x= 0 and x=16
9, which
divide the domain of f(x) into two intervals: (0,16
9) and (16
9,∞).
Since f0(1) <0 and f0(16) >0, we conclude that f0(x)<0 in (0,16
9) and f0(x)>0 in (16
9,∞).
Thus f(0) is a local maximum value, and f(16
9) is a local minimum value of f(x) in its domain.
Example 3 Find all critical points of f(x) = sin xcos xon [0,2π], and determine whether each
represents a local maximum, local minimum, or neither. (Page 267, #17)
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