200 150 100 50 Need Help? Read It Video Example 4 5 EXAMPLE 3 To illustrate the Mean Value Theorem with a specific function, let's consider f(x) = x³ = x, a = 0, b = 5. Since f is a polynomial, it is continuous and differentiable for all x, so it is certainly continuous on [0, 5] and differentiable on (0, 5). Therefore, by the Mean Value Theorem, there is a number c in (0, 5) such that f(5) f(0) = f'(c)(5-0). Now f(5) 120 = which gives = C = 2.89 secant line. ✓, f(0) = 0 X = f'(c)(5) = 125 15 X C ✓, and f'(x) = 3x² - 1 3c²1 ✓ )5 = X, that is, c = + 2.89 ✓, so this equation becomes x, X. But c must be in (0, 5), so The figure illustrates this calculation: The tangent line at this value of c is parallel to the
200 150 100 50 Need Help? Read It Video Example 4 5 EXAMPLE 3 To illustrate the Mean Value Theorem with a specific function, let's consider f(x) = x³ = x, a = 0, b = 5. Since f is a polynomial, it is continuous and differentiable for all x, so it is certainly continuous on [0, 5] and differentiable on (0, 5). Therefore, by the Mean Value Theorem, there is a number c in (0, 5) such that f(5) f(0) = f'(c)(5-0). Now f(5) 120 = which gives = C = 2.89 secant line. ✓, f(0) = 0 X = f'(c)(5) = 125 15 X C ✓, and f'(x) = 3x² - 1 3c²1 ✓ )5 = X, that is, c = + 2.89 ✓, so this equation becomes x, X. But c must be in (0, 5), so The figure illustrates this calculation: The tangent line at this value of c is parallel to the
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
I need help solving the missing blanks in the question
![200
150
100
50
Need Help? Read
It
Video Example
5
EXAMPLE 3 To illustrate the Mean Value Theorem with a specific function, let's consider f(x) = x³ - x, a = 0,
b = 5. Since f is a polynomial, it is continuous and differentiable for all x, so it is certainly continuous on [0, 5] and
differentiable on (0, 5). Therefore, by the Mean Value Theorem, there is a number c in (0, 5) such that
f(5) f(0) = f'(c)(5-0).
Now f(5) = 120
which gives 2 =
C = 2.89
secant line.
✓, f(0) = 0
X = f'(c)(5) =
125
15
X
C
✓, and f'(x) = 3x² - 1
3c²1
✓
)5 =
X, that is, c = + 2.89
✔, so this equation becomes
X,
X. But c must be in (0, 5), so
The figure illustrates this calculation: The tangent line at this value of c is parallel to the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6c921342-6d5a-48e4-a07c-a0e61e9e1942%2Fe0edad3b-b212-4dd0-a9fa-7e744c3e2561%2Fyyhjmbu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:200
150
100
50
Need Help? Read
It
Video Example
5
EXAMPLE 3 To illustrate the Mean Value Theorem with a specific function, let's consider f(x) = x³ - x, a = 0,
b = 5. Since f is a polynomial, it is continuous and differentiable for all x, so it is certainly continuous on [0, 5] and
differentiable on (0, 5). Therefore, by the Mean Value Theorem, there is a number c in (0, 5) such that
f(5) f(0) = f'(c)(5-0).
Now f(5) = 120
which gives 2 =
C = 2.89
secant line.
✓, f(0) = 0
X = f'(c)(5) =
125
15
X
C
✓, and f'(x) = 3x² - 1
3c²1
✓
)5 =
X, that is, c = + 2.89
✔, so this equation becomes
X,
X. But c must be in (0, 5), so
The figure illustrates this calculation: The tangent line at this value of c is parallel to the
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