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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 4.2.7
Chapter 4, Problem 4.2.7

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) = x²ᐟ³, [−1, 8]

Verified step by step guidance
1
Step 1: Recall the Mean Value Theorem (MVT), which 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 the continuity of f(x) = x^(2/3) on the interval [-1, 8]. A function is continuous if it has no breaks, jumps, or holes in the interval. Since x^(2/3) is a real-valued function for all x, it is continuous on [-1, 8].
Step 3: Check the differentiability of f(x) = x^(2/3) on the open interval (-1, 8). A function is differentiable at a point if it has a defined derivative at that point. Calculate the derivative f'(x) = (2/3)x^(-1/3).
Step 4: Analyze the derivative f'(x) = (2/3)x^(-1/3) for differentiability. The derivative is undefined at x = 0 because it involves division by zero. Therefore, f(x) is not differentiable at x = 0, which lies within the interval (-1, 8).
Step 5: Conclude that since f(x) = x^(2/3) is not differentiable at x = 0, it does not satisfy the hypotheses of the Mean Value Theorem on the interval [-1, 8].

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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 helps in understanding the behavior of functions and is crucial for determining if a function meets its conditions.
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Continuity

Continuity of a function on an interval 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 closed interval [a, b]. In the context of the function f(x) = x²ᐟ³, we need to check if it is continuous on the interval [-1, 8].
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Differentiability

Differentiability refers to the existence of a derivative at each point in an interval. For the Mean Value Theorem, the function must be differentiable on the open interval (a, b). For f(x) = x²ᐟ³, we must determine if the derivative exists for all x in (-1, 8), noting that differentiability implies continuity, but not vice versa.
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Related Practice
Textbook Question

Sketch the graphs of the rational functions in Exercises 53–60.


𝓍²

y = ------------------

𝓍² ― 4

261
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Textbook Question

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫(x + 1) dx

29
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Textbook Question

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫7sin(θ/3) dθ

29
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Textbook Question

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) = {sinx / x, −π ≤ x < 0

0, x = 0

172
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Textbook Question

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫2x(1 − x⁻³) dx

30
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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.


f(x) = |x|, −1 < x < 2

218
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