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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

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.1: Inverse Functions
Problem 24E
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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
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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