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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 4.2.11
Chapter 4, Problem 4.2.11

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) = {x² − x, −2 ≤ x ≤−1
2x² − 3x − 3, −1 < x ≤ 0

Verified step by step guidance
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Step 1: Understand the Mean Value Theorem (MVT). The MVT states that if a function f is continuous on a 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 interval [-2, 0]. The function f(x) is piecewise defined with f(x) = x² - x for -2 ≤ x ≤ -1 and f(x) = 2x² - 3x - 3 for -1 < x ≤ 0. Check if f(x) is continuous at x = -1 by ensuring the left-hand limit and right-hand limit at x = -1 are equal to f(-1).
Step 3: Check differentiability of f(x) on the open interval (-2, 0). For differentiability, ensure that the derivative exists at every point in the interval. Calculate the derivative of each piece: f'(x) = 2x - 1 for -2 < x < -1 and f'(x) = 4x - 3 for -1 < x < 0. Check if the derivative is continuous at x = -1.
Step 4: Determine if the hypotheses of the MVT are satisfied. If f(x) is both continuous on [-2, 0] and differentiable on (-2, 0), then the hypotheses are satisfied. If there is a discontinuity or a point where the derivative does not exist, the hypotheses are not satisfied.
Step 5: Conclude whether the function satisfies the MVT. If the hypotheses are satisfied, there exists a c in (-2, 0) such that f'(c) = (f(0) - f(-2)) / (0 - (-2)). If not, provide the reason based on the continuity or differentiability check.

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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 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) = (f(b) - f(a)) / (b - a). This theorem connects the average rate of change of the function over the interval to the instantaneous rate of change at some point within the interval.
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Continuity

Continuity of a function on a given 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]. This ensures that the function behaves predictably and smoothly across the entire interval, allowing the theorem to hold.
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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), meaning it has a defined slope or rate of change at every point within this interval. Differentiability implies continuity, but a function can be continuous without being differentiable at certain points.
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Related Practice
Textbook Question

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


𝓍²

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

𝓍² ― 4

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

Textbook Question

56. Airplane landing path An airplane is flying at altitude H when it begins its descent to an airport runway that is at horizontal ground distance L from the airplane, as shown in the accompanying figure. Assume that the landing path of the airplane is the graph of a cubic polynomial function y = ax^3+bx^2+cx+d, where y(-L)= H and y(0)=0.

a. What is dy/dx at x = 0?

b. What is dy/dx at x = -L?

c. Use the values for dy/dx at x = 0 and x =- L together with y(0) = 0 and y(-L) = H to show that y(x)=H[2(x/L)^3+3(x/L)^2]

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Textbook Question

Absolute Extrema on Finite Closed Intervals


In Exercises 21–36, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.


g(x) = √(4 − x²), −2 ≤ x ≤ 1

Textbook Question

Roots (Zeros)


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


g(t) = √t + √(1 + t) − 4, (0, ∞)

Textbook Question

Identifying Extrema


In Exercises 19–40:


a. Find the open intervals on which the function is increasing and those on which it is decreasing.


b. Identify the function’s local extreme values, if any, saying where they occur.


g(x) = x√8 − x²