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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.2.12
Chapter 4, Problem 4.2.12

Checking the Mean Value Theorem


Which of the functions in Exercises 7–12 satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers.


f(x) = {2x − 3, 0 ≤ x ≤ 2
6x − x² − 7, 2 < x ≤ 3

Verified step by step guidance
1
Step 1: Understand the Mean Value Theorem (MVT). The MVT states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in (a, b) such that f'(c) = (f(b) - f(a)) / (b - a).
Step 2: Check continuity of f(x) on the given intervals. For the piecewise function f(x), check if it is continuous at the point where the pieces meet, which is x = 2. Evaluate the left-hand limit and the right-hand limit at x = 2 to ensure they are equal.
Step 3: Check differentiability of f(x) on the open intervals. For each piece of the function, determine if the function is differentiable. For the first piece, f(x) = 2x - 3, check differentiability on (0, 2]. For the second piece, f(x) = 6x - x² - 7, check differentiability on (2, 3).
Step 4: Verify the differentiability at x = 2. Since the function is piecewise, ensure that the derivative from the left and the derivative from the right at x = 2 are equal. Calculate the derivatives of each piece and compare them at x = 2.
Step 5: Conclude whether the function satisfies the hypotheses of the MVT. If the function is both continuous on [0, 3] and differentiable on (0, 3), then it satisfies the MVT. Otherwise, identify which condition fails and explain why the MVT does not apply.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Mean Value Theorem

The Mean Value Theorem (MVT) states that for a function f that is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), there exists at least one point c in (a, b) such that f'(c) equals the average rate of change over [a, b]. This theorem is crucial for understanding how the function behaves between two points.
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Continuity

Continuity of a function on a closed interval [a, b] means that the function has no breaks, jumps, or holes in that interval. For the Mean Value Theorem to apply, the function must be continuous on the entire interval, ensuring that it can be smoothly traversed from one endpoint to the other without interruption.
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Differentiability

Differentiability refers to the existence of a derivative at each point in an open interval (a, b). A function is differentiable if it has a defined tangent at every point in the interval, meaning it is smooth without any sharp corners or cusps. Differentiability is a necessary condition for applying the Mean Value Theorem.
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Related Practice
Textbook Question

Roots (Zeros)


Show that the functions in Exercises 19–26 have exactly one zero in the given interval.


r(θ) = 2θ − cos²θ + √2, (−∞, ∞)

Textbook Question

Finding Extrema from Graphs


In Exercises 15–20, sketch the graph of each function and determine whether the function has any absolute extreme values on its domain. Explain how your answer is consistent with Theorem 1.


g(x) = {−x, 0 ≤ x < 1

x − 1, 1 ≤ x ≤ 2

Textbook Question

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.

5. y=x+sin(2x), -2π/3≤x≤2π/3

Textbook Question

Roots (Zeros)


Show that the functions in Exercises 19–26 have exactly one zero in the given interval.


g(t) = 1/(1 − t) + √(1 + t) − 3.1, (−1, 1)

Textbook Question

Root Finding

5. Use Newton's method to find the positive fourth root of 2 by solving the equation x^4 -2 = 0. Start with x_0 = 1 and find x_2.

Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


d²y/dx² = 2 − 6x; y′(0) = 4, y(0) = 1