NAME
ANSWERS
MATH 31
UNIT 5: LESSON #2
MAXIMUM AND MINIMUM VALUES
Page 1 of 6
A critical number is the value of x that makes f’(x) = 0 or f’(x) undefined
The absolute minimum of a function is the very lowest point on the graph over a given
interval of x.
The absolute maximum of a function is the very highest point on the graph over a given
interval of x.
Local minimums and maximums are turning points on the graph. They are at the values of
x for which f’(x) = 0.
Absolute maximum 5
Local maximums 2 and 3
Local minimum 1
solute and local minimum -2
Procedure for Finding the Absolute Maximum and Minimum Values of a
Continuous Function on a Closed Interval [a, b]
1. Find the values of f (x) where f ' (x) = 0
2. Find the values of f (a) and f (b).
3. The largest of the values from above is the absolute maximum value
4. The smallest of these values if the absolute minimum value
U5 L2ANS Max and Min Values .docx
NAME
ANSWERS
MATH 31
UNIT 5: LESSON #2
MAXIMUM AND MINIMUM VALUES
Page 2 of 6
EXAMPLE 1 f(x) = x2 – 2x – 8
1. Find the derivative
f’(x)= 2x – 2
2. Set f’(x) = 0 and solve for x to determine the critical number.
2x – 2 = 0
x= 1
critical number
f(1) = -9
3. Graph the function f(x) = x2 – 2x – 8 over the domain –2 < x < 3
x
-2
-1
0
1
2
3
U5 L2ANS Max and Min Values .docx
f(x)
0
-5
-8
-9
-8
-5
absolute max value is 0
absolute min value is -9
local min value is -9
NAME
ANSWERS
MATH 31
UNIT 5: LESSON #2
MAXIMUM AND MINIMUM VALUES
Page 3 of 6
EXAMPLE 2
f(x) = x3 – 9x
1. Find f’(x)
f’(x)= 3x2 – 9
2. Set f’(x) = 0 and solve for x to determine the critical values.
𝟑𝒙𝟐 − 𝟗 = 𝟎
𝟑𝒙𝟐 = 𝟗
𝒙𝟐 = 𝟑
𝒙 = ±√𝟑 = ±𝟏. 𝟕𝟑
Critical Values
𝟑
𝒇(√𝟑) = (√𝟑) − 𝟗√𝟑 = 𝟑√𝟑 − 𝟗√𝟑 = −𝟔√𝟑 = −𝟏𝟎. 𝟒
𝟑
𝒇(−√𝟑) = (−√𝟑) − 𝟗(−√𝟑) = −𝟑√𝟑 + 𝟗√𝟑 = 𝟔√𝟑 = 𝟏𝟎. 𝟒
3. Graph the function f(x) = x3 – 9x over the domain –4 < x < 3
absolute max value is 10.4
local max value is 10.4
(−𝟏. 𝟕, 𝟏𝟎. 𝟒)
(𝟑, 𝟎)
(𝟏. 𝟕, −𝟏𝟎. 𝟒)
local min value is –10.4
(−𝟒, −𝟐𝟖)
absolute min value is –28
U5 L2ANS Max and Min Values .docx
MATH 31
UNIT 5: LESSON #2
MAXIMUM AND MINIMUM VALUES
NAME ANSWERS
Page 4 of 6
ASSIGNMENT QUESTIONS
1. Find the critical values for each function then sketch the graph for the given interval
and use it to state the absolute and local maximum and minimum values.
a) f(x) = x2 + 1, –2 < x < 1
f ′(x) = 2x
x=0
x
-2
0
1
b) f(x) = x2 + 1, –1 < x < 2
x
f ′(x) = 2x
-1
0
2
x=0
f(x)
5
1
2
absolute maximum is 5
absolute maximum is 5
absolute and local minimum is 1
absolute and local minimum is 1
c) f(x) = – x2 + 4, –2 < x < 3
f ′(x) = -2x
x=0
f(x)
2
1
5
d) f(x) = x4 – 8x2 + 6, –3 < x < 3
x
-2
0
3
f(x)
0
4
-5
f ′(x) = 4x3 – 16x
4x3 – 16x = 0
4x(x2 – 4) = 4x (x – 2)(x + 2) = 0
x = -2 or
x=2
Vertical scale is 3
absolute and local maximum is 4
absolute maximum is 15
absolute minimum is -5
local minimums are -10
local maximum is 6
U5 L2ANS Max and Min Values .docx
x
-3
-2
0
2
3
f(x)
15
-10
6
-10
15
NAME
ANSWERS
MATH 31
UNIT 5: LESSON #2
MAXIMUM AND MINIMUM VALUES
Page 5 of 6
(𝟎, 𝟏)
2. Find the absolute and local maximum and minimum values.
a) f(x) = 2x2 – 8x + 1, 0 < x < 3
f ′ (x) = 4x – 8
4x – 8 = 0
x=2
x
0
2
3
f(x)
1
-7
-5
critical value is x = 2
(𝟑, −𝟓)
f(0) = 2(0)2 – 8(0) + 1 = 1
f(critical value) = f(2) = 2(2)2 – 8(2) + 1 = –7
f(3) = 2(3)2 – 8(3) + 1 = –5
absolute max value is 1
no local max value
(𝟐, −𝟕)
absolute min value is –7
local min value is –7
(𝟒, 𝟐𝟎)
(𝟏, 𝟏𝟕)
𝟏𝟔
b) 𝒈(𝒙) = 𝒙 +
,𝟏 ≤ 𝒙 ≤ 𝟒
𝒙
𝟐
𝟏𝟔
𝒈′ (𝒙) = 𝟐𝒙 − 𝟐
𝒙
𝟏𝟔
𝟎 = 𝟐𝒙 − 𝟐
𝒙
𝟏𝟔
𝟐𝒙 = 𝟐
𝒙
𝒙𝟑 = 𝟖
𝒙=𝟐
x
1
2
4
f(x)
17
12
20
critical value is 2
g(1) = 17
g(2) = 12
local and absolute minimum is 12
g(4) = 20
absolute maximum is 20
U5 L2ANS Max and Min Values .docx
(𝟐, 𝟏𝟐)
NAME
MATH 31
UNIT 5: LESSON #2
MAXIMUM AND MINIMUM VALUES
Page 6 of 6
ANSWERS
c)
𝟐
𝒇(𝒙) = 𝟑𝒙𝟑 − 𝟐𝒙,
−
′ (𝒙)
𝒇
(−𝟏, 𝟓)
−𝟏≤𝒙≤𝟖
(𝟏, 𝟏)
𝟏
𝟑
= 𝟐𝒙 − 𝟐
𝟐
𝟎= 𝟑 −𝟐
√𝒙
𝟐
𝟐=𝟑
√𝒙
𝟑
𝒙
=
𝟏
√
𝒙=𝟏
x
-1
1
8
f(x)
5
1
-4
(𝟖, −𝟒)
critical value is 1
d)
f(-1) = 5
absolute maximum value is 5
f(1) = 1
local maximum value is 1; there is no local minimum value
f(8) = –4
absolute minimum value is –4
(−𝟐, 𝟏𝟏)
𝒇(𝒙) = 𝒙𝟑 − 𝟏𝟐𝒙 − 𝟓, −𝟒. 𝟏 ≤ 𝒙 ≤ 𝟒. 𝟏
f ′(x) = 3x2 – 12
3x2
x
-4.1
-2
2
4.1
– 12 = 0
x2 = 4
x=+2
(𝟒. 𝟏, 𝟏𝟒. 𝟕)
f(x)
–24.7
11
–21
14.7
critical values are –2 and 2
f(–4.1) = –24.7
absolute minimum value is –24.7
f(–2) = 11
local maximum value is 11
f(2) = –21
local minimum value is –21
f(4.1) = 14.7
absolute maximum value is 14.7
U5 L2ANS Max and Min Values .docx
(−𝟐, −𝟐𝟏)
(−𝟒. 𝟏, −𝟐𝟒. 𝟕)