39. y = x² +1 x - 4 [5,6]

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.3: Algebraic Expressions
Problem 44E
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Question
Number 39
In Exercises 29–58, find the minimum and maximum value of the func-
tion on the given interval by comparing values at the critical points and
endpoints.
Transcribed Image Text:In Exercises 29–58, find the minimum and maximum value of the func- tion on the given interval by comparing values at the critical points and endpoints.
39.
=
x² +1
X- - 4
41. y = x -
4x
x+1
55. y =
[5,6]
X
9
42. y = 2√√x² +1−x, [0, 2]
43. y = (2 + x)√√2+ (2 − x)², [0, 2]
44. y = √1+x²–2x, [0, 1]
[0, 4]
45. y = √√√x + x²-2√x,
47. y sin x cos x, [0, 1]
49. y = √20 sec 0, [0, 1]
50. y = cos 0 + sin 0,
[0, 2л]
X
[0, 3]
51. y = 0 - 2 sin 0,
[0, 2π]
52. y = 4 sin³ 0 - 3 cos²0, [0, 2π]
53. y = tan x-2x, [0, 1]
ln x
[1, 3]
40. y
57. y = 3ete²x, [−1,¹]
=
1- x
x² + 3x²
[1,4]
46. y = (t t²)1/3, [1,2]
48. y = x + sinx, [0, 2π]
54. y = xe-*, [0, 2]
56. y = 5 tan
58. y = x³ - 24 ln x,
-1
x -x, [1, 5]
[1/2,3]
Transcribed Image Text:39. = x² +1 X- - 4 41. y = x - 4x x+1 55. y = [5,6] X 9 42. y = 2√√x² +1−x, [0, 2] 43. y = (2 + x)√√2+ (2 − x)², [0, 2] 44. y = √1+x²–2x, [0, 1] [0, 4] 45. y = √√√x + x²-2√x, 47. y sin x cos x, [0, 1] 49. y = √20 sec 0, [0, 1] 50. y = cos 0 + sin 0, [0, 2л] X [0, 3] 51. y = 0 - 2 sin 0, [0, 2π] 52. y = 4 sin³ 0 - 3 cos²0, [0, 2π] 53. y = tan x-2x, [0, 1] ln x [1, 3] 40. y 57. y = 3ete²x, [−1,¹] = 1- x x² + 3x² [1,4] 46. y = (t t²)1/3, [1,2] 48. y = x + sinx, [0, 2π] 54. y = xe-*, [0, 2] 56. y = 5 tan 58. y = x³ - 24 ln x, -1 x -x, [1, 5] [1/2,3]
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