Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
Maximum and Minimum Values
DEFINITION: A function f has an absolute maximum (or global maximum) at c if
f (c) ≥ f (x) for all x in D, where D is the domain of f. The number f (c) is called the
maximum value of f on D. Similarly, f has an absolute minimum (or global minimum)
at c if f (c) ≤ f (x) for all x in D and the number f (c) is called the minimum value of f on
D. The maximum and minimum values of f are called the extreme values of f.
DEFINITION: A function f has a local maximum (or relative maximum) at c if f (c) ≥
f (x) when x is near c. [This means that f (c) ≥ f (x) for all x in some open interval containing
c.] Similarly, f has a local minimum (or relative minimum) at c if f (c) ≤ f (x) when x is
near c.
EXAMPLES:
1. The function f (x) = x2 has the absolute (and local) minimum at x = 0 and has no absolute
or local maximum.
2. The function f (x) = x2 , x ∈ (−∞, 0) ∪ (0, ∞), has no absolute or local minimum and no
absolute or local maximum.
f (x) = x2 , x ∈ (−∞, 0) ∪ (0, ∞)
3. The functions f (x) = x, x3 , x5 , etc. have no absolute or local minima and no absolute or
local maxima.
1
Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
4. The functions f (x) = sin x, cos x have infinitely many absolute and local minima and
maxima.
5. The functions f (x) = sec x, csc x have infinitely many local minima and maxima and no
absolute minima and maxima.
6. The functions f (x) = tan x, cot x have no absolute or local minima and no absolute or local
maxima.
2
Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
7. The function f (x) = x4 + x3 − 11 x2 − 9 x + 18 = (x − 3)(x − 1)(x + 2)(x + 3) has the absolute
minimum at x ≈ 2.2 and has no absolute maximum. It has two local minima at x ≈ −2.6 and
x ≈ 2.2 and the local maximum at x ≈ −0.4.
8. The function f (x) = x, 1 ≤ x ≤ 2, has the absolute minimum at x = 1 and absolute
maximum at x = 2. It has no local maximum or minimum.
9. The function f (x) = x, 1 < x < 2, has no absolute or local minimum and no absolute or
local maximum.
10. The function f (x) = x has no absolute or local minimum and no absolute or local maximum.
−2
2
2
2
1
1
1
−1
−1
1
2
−2
−2
−1
−1
1
2
−2
−2
−1
−1
1
2
−2
11. The function f (x) = 1 has the absolute (and local) minimum and absolute (and local)
maximum at any point on the number line.
2
1
−2
−1
−1
1
−2
3
2
Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
THEOREM (The Extreme Value Theorem): If f is continuous on a closed interval [a, b], then f
attains an absolute maximum value f (c) and an absolute minimum value f (d) at some numbers
c and d in [a, b].
THEOREM (Fermat’s Theorem): If f has a local maximum or minimum at c, and if f ′ (c)
exists, then f ′ (c) = 0.
REMARK 1: The converse of this theorem is not true. In other words, when f ′ (c) = 0, f does
not necessarily have a local maximum or minimum. For example, if f (x) = x3 , then f ′ (x) = 3x2
equals 0 at x = 0, but x = 0 is not a point of a local minimum or maximum.
REMARK 2: Sometimes f ′ (c) does not exist, but x = c is a point of a local maximum or
minimum. For example, if f (x) = |x|, then f ′ (0) does not exist. But f (x) has its local (and
absolute) minimum at x = 0.
4
Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
DEFINITION: A critical number of a function f is a number c in the domain of f such that
either f ′ (c) = 0 or f ′ (c) does not exist.
REMARK: From Fermat’s Theorem it follows that if f has a local maximum or minimum at
c, then c is a critical number of f.
EXAMPLES:
5
(a) If f (x) = 2x2 + 5x − 1, then f ′ (x) = 4x + 5. Hence the only critical number of f is x = − .
4
√
2
2
3
(b) If f (x) = x2 , then f ′ (x) = x−1/3 = √
. Hence the only critical number of f is x = 0.
3
33x
(c) If f (x) =
1
1
, then f ′ (x) = − 2 . Since x = 0 is not in the domain, f has no critical numbers.
x
x
2
2
1
1
3
2
1
−3 −2 −1
−1
1
−3 −2 −1
1
2
3
−3 −2 −1
−1
−2
−2
−3
−3
−4
EXAMPLES:
(a) Find the critical numbers of f (x) = 2x3 − 9x2 + 12x − 5.
√
3
(b) Find the critical numbers of f (x) = 2x + 3 x2 .
5
1
2
3
Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
EXAMPLES:
3
(a) Find the critical numbers of f (x) = 2x3 − 9x2 + 12x − 5.
2
Solution: We have
1
2
′
f (x) = 6x − 18x + 12
= 6(x2 − 3x + 2)
−1
= 6(x − 1)(x − 2)
−2
1
2
3
thus f ′ (x) = 0 at x = 1 and x = 2. Since f ′ (x) exists everywhere, x = 1 and x = 2 are the only
critical numbers.
√
3
(b) Find the critical numbers of f (x) = 2x + 3 x2 .
Solution: We have
f ′ (x) = (2x + 3x2/3 )′
= 2x′ + 3(x2/3 )′
2
= 2 · 1 + 3 · x2/3−1
3
√
√
√
2
2( 3 x + 1)
23x
2
23x+2
2
−1/3
√
+√
=
= √
+√
= √
= 2 + 2x
=
3
3
3
3
3
1
x
x
x
x
x
or
= 2 + 2x
−1/3
n
= 2x
−1/3
·x
1/3
+ 2x
−1/3
In short,
′
f (x) = 2 + 2x
−1/3
= 2x
−1/3
(x
1/3
· 1 = 2x
−1/3
(x
1/3
√
2( 3 x + 1)
√
+ 1) =
3
x
√
2( 3 x + 1)
√
+ 1) =
3
x
o
√
′
3
It
follows
that
f
(x)
equals
0
when
x + 1 = 0, which is at x = −1; f ′ (x) does not exist when
√
3
x = 0, which is at x = 0. Thus the critical numbers are −1 and 0.
2
1
−5 −4 −3 −2 −1
−1
1
EXAMPLE: Find the critical numbers of f (x) = x1/3 (2 − x).
6
Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
EXAMPLE: Find the critical numbers of f (x) = x1/3 (2 − x).
Solution: We have
f ′ (x) = [x1/3 (2 − x)]′
1
= (x1/3 )′ (2 − x) + x1/3 (2 − x)′
1
= x−2/3 (2 − x) + x1/3 · (−1)
3
=
1
2
3
4
−2
−3
2−x
= 2/3 − x1/3
3x
−2 −1
−1
2−x
3x
2 − x − 3x
2 − 4x
2 − x x1/3 · 3x2/3
−
= 2/3 − 2/3 =
=
2/3
2/3
2/3
3x
3x
3x
3x
3x
3x2/3
=
2(1 − 2x)
3x2/3
In short,
f ′ (x) = (x1/3 )′ (2 − x) + x1/3 (2 − x)′ =
2−x
2(1 − 2x)
− x1/3 =
2/3
3x
3x2/3
Here is an other way to get the same result:
f ′ (x) = [x1/3 (2 − x)]′
= (2x1/3 − x4/3 )′
= 2(x1/3 )′ − (x4/3 )′
1
4
= 2 · x1/3−1 − x4/3−1
3
3
2
4
= x−2/3 − x1/3 =
3
3
4x1/3
2
4x1/3 · x2/3
2
4x
2 − 4x
2
−
=
−
= 2/3 − 2/3 =
2/3
2/3
2/3
3x
3
3x
3·x
3x
3x
3x2/3
=
or
4
2
= x−2/3 − x1/3 =
3
3
2
2
2 −2/3
x
· 1 − x−2/3 · 2x = x−2/3 (1 − 2x)
3
3
3
=
2(1 − 2x)
3x2/3
2(1 − 2x)
3x2/3
In short,
2
4
2
2(1 − 2x)
f ′ (x) = (2x1/3 − x4/3 )′ = x−2/3 − x1/3 = x−2/3 (1 − 2x) =
.
3
3
3
3x2/3
1
It follows that f ′ (x) equals 0 when 1 − 2x = 0, which is at x = ; f ′ (x) does not exist when
2
1
2/3
x = 0, which is at x = 0. Thus the critical numbers are and 0.
2
7
Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
THE CLOSED INTERVAL METHOD: To find the absolute maximum and minimum values
of a continuous function f on a closed interval [a, b]:
1. Find the values of f at the critical numbers of f in (a, b).
2. Find the values of f at the endpoints of the interval.
3. The largest of the values from Step 1 and 2 is the absolute maximum value; the smallest
value of these values is the absolute minimum value.
EXAMPLE: Find the absolute maximum and minimum values of f (x) = 2x3 − 15x2 + 36x on
the interval [1, 5] and determine where these values occur.
Solution:
Step 1: Since
f ′ (x) = 6x2 − 30x + 36 = 6(x2 − 5x + 6) = 6(x − 2)(x − 3)
there are two critical numbers x = 2 and x = 3.
Step 2: We now evaluate f at these critical numbers and at the endpoints x = 1 and x = 5.
We have
f (1) = 23, f (2) = 28, f (3) = 27, f (5) = 55
Step 3: The largest value is 55 and the smallest value is 23. Therefore the absolute maximum
of f on [1, 5] is 55, occurring at x = 5 and the absolute minimum of f on [1, 5] is 23, occurring
at x = 1.
EXAMPLES:
(a) Find the absolute maximum and minimum values of f (x) = 2x3 − 15x2 + 24x + 2 on [0, 2]
and determine where these values occur.
(b) Find the absolute maximum and minimum values of f (x) = 6x4/3 − 3x1/3 on the interval
[−1, 1] and determine where these values occur.
8
Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
EXAMPLES:
(a) Find the absolute maximum and minimum values of f (x) = 2x3 − 15x2 + 24x + 2 on [0, 2]
and determine where these values occur.
Solution:
Step 1: Since
f ′ (x) = 6x2 − 30x + 24 = 6(x2 − 5x + 4) = 6(x − 4)(x − 1)
there are two critical numbers x = 1 and x = 4.
Step 2: Since x = 4 is not from [0, 2], we evaluate f only at x = 1 and at the endpoints x = 0
and x = 2. We have
f (0) = 2, f (1) = 13, f (2) = 6
Step 3: The largest value is 13 and the smallest value is 2. Therefore the absolute maximum
of f on [0, 2] is 13, occurring at x = 1 and the absolute minimum of f on [0, 2] is 2, occurring
at x = 0.
(b) Find the absolute maximum and minimum values of f (x) = 6x4/3 − 3x1/3 on the interval
[−1, 1] and determine where these values occur.
Solution:
Step 1: Since
4
1
f ′ (x) = (6x4/3 − 3x1/3 )′ = 6(x4/3 )′ − 3(x1/3 )′ = 6 · x4/3−1 − 3 · x1/3−1
3
3
= 8x1/3 − x−2/3
= 8x−2/3 · x − 1 · x−2/3
= x−2/3 (8x − 1)
=
8x − 1
x2/3
1
there are two critical numbers x = 0 and x = .
8
Step 2: We now evaluate f at these critical numbers and at the endpoints x = −1 and x = 1.
We have
9
1
f (−1) = 9, f (0) = 0, f
= − , f (1) = 3
8
8
9
Step 3: The largest value is 9 and the smallest value is − . Therefore the absolute maximum
8
9
of f on [−1, 1] is 9, occurring at x = −1 and the absolute minimum of f on [−1, 1] is − ,
8
1
occurring at x = .
8
9
Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
Appendix
EXAMPLE: Find the absolute maximum and minimum values of f (x) = 2x + 3 on [−1, 4] and
determine where these values occur.
Solution:
Step 1: Since
f ′ (x) = (2x + 3)′ = (2x)′ − 3′ = 2(x)′ − 3′ = 2 · 1 − 0 = 2
there are no critical numbers.
Step 2: Since there are no critical numbers, we evaluate f at the endpoints x = −1 and x = 4
only. We have
f (−1) = 2(−1) + 3 = −2 + 3 = 1
f (4) = 2(4) + 3 = 8 + 3 = 11
Step 3: The largest value is 11 and the smallest value is 1. Therefore the absolute maximum of
f on [−1, 4] is 11, occurring at x = 4 and the absolute minimum of f on [−1, 4] is 1, occurring
at x = −1.
EXAMPLE: Find the absolute maximum and minimum values of f (x) = 2x2 − 8x + 1 on [−1, 3]
and determine where these values occur.
Solution:
Step 1: Since
f ′ (x) = (2x2 − 8x + 1)′ = (2x2 )′ − (8x)′ + 1′ = 2(x2 )′ − 8(x)′ + 1′ = 2 · 2x − 8 · 1 + 0 = 4x − 8
the number at which f ′ is zero is 2. Therefore x = 2 is the critical number.
Step 2: We evaluate f at the critical number x = 2 and at the endpoints x = −1, 3. We have
f (2) = 2 · 22 − 8 · 2 + 1 = 2 · 4 − 8 · 2 + 1 = 8 − 16 + 1 = −7
f (−1) = 2(−1)2 − 8(−1) + 1 = 2 + 8 + 1 = 11
f (3) = 2 · 32 − 8 · 3 + 1 = 2 · 9 − 8 · 3 + 1 = 18 − 24 + 1 = −5
Step 3: The largest value is 11 and the smallest value is −7. Therefore the absolute maximum
of f on [−1, 3] is 11, occurring at x = −1 and the absolute minimum of f on [−1, 3] is −7,
occurring at x = 2.
EXAMPLE: Find the absolute maximum and minimum values of f (x) = 5 − x3 on [−2, 1] and
determine where these values occur.
10
Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
EXAMPLE: Find the absolute maximum and minimum values of f (x) = 5 − x3 on [−2, 1] and
determine where these values occur.
Solution:
Step 1: Since
f ′ (x) = (5 − x3 )′ = 5′ − (x3 )′ = 0 − 3x2 = −3x2
the critical number is x = 0.
Step 2: We evaluate f at the critical number x = 0 and at the endpoints x = −2, 1. We have
f (0) = 5 − 03 = 5 − 0 = 5
f (−2) = 5 − (−2)3 = 5 − (−8) = 5 + 8 = 13
f (1) = 5 − 13 = 5 − 1 = 4
Step 3: The largest value is 13 and the smallest value is 4. Therefore the absolute maximum of
f on [−2, 1] is 13, occurring at x = −2 and the absolute minimum of f on [−2, 1] is 4, occurring
at x = 1.
EXAMPLE: Find the absolute maximum and minimum values of f (x) = x3 + 3x2 − 1 on [−3, 2]
and determine where these values occur.
Solution:
Step 1: Since
f ′ (x) = (x3 + 3x2 − 1)′ = (x3 )′ + (3x2 )′ − 1′ = (x3 )′ + 3(x2 )′ − 1′ = 3x2 + 3 · 2x − 0 = 3x2 + 6x
= 3x(x + 2)
the critical numbers are x = 0 and x = −2.
Step 2: We evaluate f at the critical numbers x = 0, −2 and at the endpoints x = −3, 2. We
have
f (0) = 03 + 3 · 02 − 1 = 0 + 0 − 1 = −1
f (−2) = (−2)3 + 3 · (−2)2 − 1 = −8 + 12 − 1 = 3
f (−3) = (−3)3 + 3 · (−3)2 − 1 = −27 + 27 − 1 = −1
f (2) = 23 + 3 · 22 − 1 = 8 + 12 − 1 = 19
Step 3: The largest value is 19 and the smallest value is −1. Therefore the absolute maximum
of f on [−3, 2] is 19, occurring at x = 2 and the absolute minimum of f on [−3, 2] is −1,
occurring at x = 0 and x = −3.
11
Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
EXAMPLE: Find the absolute maximum and minimum values of f (x) = 3x4 + 4x3 − 36x2 on
[−2, 3] and determine where these values occur.
Solution:
Step 1: Since
f ′ (x) = (3x4 + 4x3 − 36x2 )′ = (3x4 )′ + (4x3 )′ − (36x2 )′ = 3(x4 )′ + 4(x3 )′ − 36(x2 )′
= 3 · 4x3 + 4 · 3x2 − 36 · 2x
= 12x3 + 12x2 − 72x
= 12x(x2 + x − 6)
= 12x(x − 2)(x + 3)
the critical numbers are x = 0, x = 2, and x = −3.
Step 2: Since −3 6∈ [−2, 3], we evaluate f only at the critical numbers x = 0, 2 and at the
endpoints x = −2, 3. We have
f (0) = 3 · 04 + 4 · 03 − 36 · 02 = 0 + 0 − 0 = 0
f (2) = 3 · 24 + 4 · 23 − 36 · 22 = 48 + 32 − 144 = −64
f (−2) = 3 · (−2)4 + 4 · (−2)3 − 36 · (−2)2 = 48 − 32 − 144 = −128
f (3) = 3 · 34 + 4 · 33 − 36 · 32 = 243 + 108 − 324 = 27
Step 3: The largest value is 27 and the smallest value is −128. Therefore the absolute maximum
of f on [−2, 3] is 27, occurring at x = 3 and the absolute minimum of f on [−2, 3] is −128,
occurring at x = −2.
1
EXAMPLE: Find the absolute maximum and minimum values of f (x) = − 2 on [0.4, 5] and
x
determine where these values occur.
Solution:
Step 1: We have
′
f (x) =
1
− 2
x
′
=−
1
x2
′
= − x−2
′
= −(−2)x−2−1 = 2x−3 =
2
x3
Note that there are no numbers at which f ′ is zero. The number at which f ′ does not exist is
x = 0, but this number is not from the domain of f. Therefore f does not have critical numbers.
Step 2: Since f does not have critical numbers, we evaluate it only at the endpoints x = 0.4
and x = 5. We have
1
1
1
1
f (0.4) = −
=−
= −6.25
f (5) = − 2 = − = −0.04
2
(0.4)
0.16
5
25
Step 3: The largest value is −0.04 and the smallest value is −6.25. Therefore the absolute
maximum of f on [0.4, 5] is −0.04, occurring at x = 5 and the absolute minimum of f on [0.4, 5]
is −6.25, occurring at x = 0.4.
12
Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
12
on [2, 4]
EXAMPLE: Find the absolute maximum and minimum values of f (x) = 20 − 3x −
x
and determine where these values occur.
Solution:
Step 1: We have
′
f (x) =
12
20 − 3x −
x
′
= 20 − 3x − 12x−1
′
= 20′ − (3x)′ − 12x−1
′
= 20′ − 3 · x′ − 12 x−1
′
= 0 − 3 · 1 − 12(−1)x−1−1
= −3 + 12x−2
Since
−3 + 12x−2 = −3 + 12 ·
1
12
−3x2 12
−3x2 + 12
=
−3
+
=
+
=
x2
x2
x2
x2
x2
=
−3(x2 − 4)
x2
=
−3(x2 − 22 )
x2
=
−3(x − 2)(x + 2)
x2
we have
3(x − 2)(x + 2)
x2
The numbers at which f ′ is zero are x = 2 and x = −2. The number at which f ′ does not exist
is x = 0. But f is not defined at x = 0, therefore the critical numbers of f are x = −2 and
x = 2 only.
f ′ (x) = −
Step 2: Since −2 6∈ [2, 4], we evaluate f only at the critical number x = 2 (which is also one
of the endpoints) and at the second endpoint x = 4. We have
f (2) = 20 − 3 · 2 −
12
= 20 − 6 − 6 = 8
2
f (4) = 20 − 3 · 4 −
12
= 20 − 12 − 3 = 5
4
Step 3: The largest value is 8 and the smallest value is 5. Therefore the absolute maximum of
f on [2, 4] is 8, occurring at x = 2 and the absolute minimum of f on [2, 4] is 5, occurring at
x = 4.
13
Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
√
EXAMPLE: Find the absolute maximum and minimum values of f (x) = 2 3 x on [−8, 1] and
determine where these values occur.
Solution:
Step 1: We have
f ′ (x) = 2x1/3
′
= 2 x1/3
′
2
2
1
= 2 · x1/3−1 = x−2/3 = 2/3
3
3
3x
Note that there are no numbers at which f ′ is zero. The number at which f ′ does not exist is
x = 0, so the critical number is x = 0.
Step 2: We evaluate f at the critical number x = 0 and at the endpoints x = −8 and x = 1.
We have
√
3
f (0) = 2 0 = 2 · 0 = 0
√
f (−8) = 2 3 −8 = 2 · (−2) = −4
√
3
f (1) = 2 1 = 2 · 1 = 2
Step 3: The largest value is 2 and the smallest value is −4. Therefore the absolute maximum of
f on [−8, 1] is 2, occurring at x = 1 and the absolute minimum of f on [−8, 1] is −4, occurring
at x = −8.
EXAMPLE: Find the absolute maximum and minimum values of f (x) = −5x2/3 on [−1, 1] and
determine where these values occur.
Solution:
Step 1: We have
f ′ (x) = −5x2/3
′
= −5 x2/3
′
10
10
2
= −5 · x2/3−1 = − x−1/3 = − 1/3
3
3
3x
Note that there are no numbers at which f ′ is zero. The number at which f ′ does not exist is
x = 0, so the critical number is x = 0.
Step 2: We evaluate f at the critical number x = 0 and at the endpoints x = −1 and x = 1.
We have
f (0) = −5(0)2/3 = −5 · 0 = 0
1/3
f (−1) = −5(−1)2/3 = −5 (−1)2
= −5(1)1/3 = −5 · 1 = −5
f (1) = −5(1)2/3 = −5 · 1 = −5
Step 3: The largest value is 0 and the smallest value is −5. Therefore the absolute maximum of
f on [−1, 1] is 0, occurring at x = 0 and the absolute minimum of f on [−1, 1] is −5, occurring
at x = ±1.
EXAMPLE: Find the absolute maximum and minimum values of f (x) = x4/3 on [−27, 27] and
determine where these values occur.
14
Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
EXAMPLE: Find the absolute maximum and minimum values of f (x) = x4/3 on [−27, 27] and
determine where these values occur.
Solution:
Step 1: We have
4
4
= x4/3−1 = x1/3
3
3
Note that there are no numbers at which f ′ does not exist. The number at which f ′ is zero is
x = 0, so the critical number is x = 0.
f ′ (x) = x4/3
′
Step 2: We evaluate f at the critical number x = 0 and at the endpoints x = −27 and x = 27.
We have
f (0) = 04/3 = 0
4
f (−27) = (−27)4/3 = (−27)1/3 = (−3)4 = 81
4
f (27) = 274/3 = 271/3 = 34 = 81
Step 3: The largest value is 81 and the smallest value is 0. Therefore the absolute maximum
of f on [−27, 27] is 81, occurring at x = ±27 and the absolute minimum of f on [−27, 27] is 0,
occurring at x = 0.
EXAMPLE: Find the absolute maximum and minimum values of f (x) =
and determine where these values occur.
√
4 − x2 on [−2, 1]
Solution:
Step 1: We have
f ′ (x) = (4 − x2 )1/2
′
1
1
= (4 − x2 )1/2−1 · (4 − x2 )′ = (4 − x2 )−1/2 · (4′ − (x2 )′ )
2
2
1
1
= (4 − x2 )−1/2 · (0 − 2x) = (4 − x2 )−1/2 · (−2x) = −x(4 − x2 )−1/2
2
2
x
x
= −√
=−
2
1/2
(4 − x )
4 − x2
Note that the number at which f ′ is zero is x = 0 and the numbers at which f ′ does not exist
are x = ±2. Therefore the critical numbers are x = 0, −2, 2.
Step 2: Since 2 6∈ [−2, 1], we evaluate f only at the critical numbers x = 0, −2 (which is also
one of the endpoints) and at the second endpoint x = 1. We have
√
√
√
f (0) = 4 − 02 = 4 − 0 = 4 = 2
p
√
√
f (−2) = 4 − (−2)2 = 4 − 4 = 0 = 0
√
√
√
f (1) = 4 − 12 = 4 − 1 = 3
Step 3: The largest value is 2 and the smallest value is 0. Therefore the absolute maximum of
f on [−2, 1] is 2, occurring at x = 0 and the absolute minimum of f on [−2, 1] is 0, occurring
at x = −2.
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Section 4.1 Maximum and Minimum Values
2010 Kiryl Tsishchanka
π 5π
EXAMPLE: Find the absolute maximum and minimum values of f (x) = cos x on − ,
2 2
and determine where these values occur.
Solution:
Step 1: We have
f ′ (x) = − sin x
π 5π
Note that there are no numbers at which f does not exist. The numbers in − ,
at which
2 2
f ′ is zero on are x = 0, π, 2π, so the critical number is x = 0, π, 2π.
′
π 5π
Step 2: We evaluate f at the critical numbers x = 0, π, 2π and at the endpoints x = − , .
2 2
We have
f (0) = cos 0 = 1
f (π) = cos π = −1
f (2π) = cos(2π) = 1
π
π
f −
= cos −
=0
2
2
5π
5π
= cos
=0
f
2
2
Step 3:The largest
value is 1 and the smallest value is −1. Therefore the absolute
maximum
π 5π
π 5π
is 1, occurring at x = 0, 2π and the absolute minimum of f on − ,
is
of f on − ,
2 2
2 2
−1, occurring at x = π.
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