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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 4.2.6
Chapter 4, Problem 4.2.6

Checking the Mean Value Theorem


Find the value or values of c that satisfy the equation (f(b) − f(a)) / (b − a) = f′(c) in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises 1–6.


g(x) = {x³, −2 ≤ x ≤ 0
x², 0 < x ≤ 2

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1
First, understand the Mean Value Theorem (MVT). It states that for a function g(x) that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), there exists at least one c in (a, b) such that (g(b) - g(a)) / (b - a) = g'(c).
Identify the intervals and the function pieces: g(x) = x³ for -2 ≤ x ≤ 0 and g(x) = x² for 0 < x ≤ 2. The function is continuous on [-2, 2] but we need to check differentiability at x = 0.
Check differentiability at x = 0 by finding the derivatives of each piece: For x³, the derivative is g'(x) = 3x², and for x², the derivative is g'(x) = 2x. Evaluate the left-hand and right-hand derivatives at x = 0 to ensure they are equal.
Calculate g(-2) and g(2) to apply the MVT: g(-2) = (-2)³ = -8 and g(2) = (2)² = 4. Use these to find the average rate of change: (g(2) - g(-2)) / (2 - (-2)) = (4 - (-8)) / 4.
Set the average rate of change equal to the derivative g'(c) and solve for c in each interval: For -2 < c < 0, use g'(c) = 3c², and for 0 < c < 2, use g'(c) = 2c. Solve these equations to find the value(s) of c that satisfy the MVT.

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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 continuous function on a closed interval [a, b] that is differentiable on the open interval (a, b), there exists at least one point c in (a, b) such that the derivative at c equals the average rate of change over [a, b]. This is expressed as f'(c) = (f(b) - f(a)) / (b - a).
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Fundamental Theorem of Calculus Part 1

Piecewise Functions

A piecewise function is defined by different expressions over different intervals. In this problem, g(x) is defined as x³ for -2 ≤ x ≤ 0 and x² for 0 < x ≤ 2. Understanding how to evaluate and differentiate each piece within its respective interval is crucial for applying the Mean Value Theorem.
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Piecewise Functions

Differentiability and Continuity

For the Mean Value Theorem to apply, the function must be continuous on the closed interval and differentiable on the open interval. This requires checking that g(x) is continuous at x = 0, where the pieces meet, and ensuring differentiability across the entire interval, except possibly at the endpoints.
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Related Practice
Textbook Question

Finding Extreme Values

In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.


y = 𝓍³ ― 2𝓍 + 4

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) = √(x(1 − x)), [0, 1]

Textbook Question

Applications


Stopping a motorcycle The State of Illinois Cycle Rider Safety Program requires motorcycle riders to be able to brake from 30 mph (44 ft/sec) to 0 in 45 ft. What constant deceleration does it take to do that?

Textbook Question

Finding Extrema from Graphs


In Exercises 11–14, match the table with a graph.


1
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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 + ³√x) dx

Textbook Question

37. What value of a makes f(x) = x^2 +(a/x) have

a. a local minimum at x = 2?

b. a point of inflection at x = 1?