To illustrate the Mean Value Theorem with a specific function, let's consider f(x) = x - x, a = 0, b = 8. Since f is a polynomial, it is continuous and differentiable for all x, so it is certainly continuous on [0, 8] and differentiable on (0, 8). Therefore, by the Mean Value Theorem, there is a number c in (0, 8) such that f(8) – f(0) = f'(c)(8 - 0). Now f(8) = , f(0) = , and f'(x) = so this equation becomes = f'(c)(8) = (8) = which gives c2 = , that is, c = But c must be in (0, 8), so c =

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.3: Zeros Of Polynomials
Problem 65E
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To illustrate the Mean Value Theorem with a specific function, let's consider f(x) = x³ - x, a = 0, b = 8. Since f is a polynomial, it is continuous and differentiable for all x, so it is certainly continuous on
[0, 8] and differentiable on (0, 8). Therefore, by the Mean Value Theorem, there is a number c in (0, 8) such that
f(8) – f(0) = f'(c)(8 – 0).
Now f(8)
f(0) =
and f'(x) =
, so this equation becomes
=
f'(c)(8) =
|(8)
which gives c
that is, c = ±
But c must be in (0, 8), so c =
The following figure illustrates the calculation that the tangent line at this value of c is parallel to the secant line.
y
700
600
500
400
300
200
100
8
Transcribed Image Text:To illustrate the Mean Value Theorem with a specific function, let's consider f(x) = x³ - x, a = 0, b = 8. Since f is a polynomial, it is continuous and differentiable for all x, so it is certainly continuous on [0, 8] and differentiable on (0, 8). Therefore, by the Mean Value Theorem, there is a number c in (0, 8) such that f(8) – f(0) = f'(c)(8 – 0). Now f(8) f(0) = and f'(x) = , so this equation becomes = f'(c)(8) = |(8) which gives c that is, c = ± But c must be in (0, 8), so c = The following figure illustrates the calculation that the tangent line at this value of c is parallel to the secant line. y 700 600 500 400 300 200 100 8
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