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.
∫(2x³ − 5x + 7) dx
Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
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 guidance1
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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Fundamental Theorem of Calculus Part 1
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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05:34Intro to Continuity
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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05:53Finding Differentials
Related Practice
Textbook Question
Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.
74. y' = (x² - 2x)(x - 5)²
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 + 𝓍² ― 8𝓍 + 5
Textbook Question
109. Suppose the derivative of the function y = f(x) is
y'=(x-1)^2(x-2).
At what points, if any, does the graph of f have a local minimum, local maximum, or
point of inflection? (Hint: Draw the sign pattern for y'.)
Textbook Question
"Roots (Zeros) Show that the functions in Exercises 19–26 have exactly one zero
Textbook Question
Roots (Zeros)
Show that the functions in Exercises 19–26 have exactly one zero in the given interval.
f(x) = x⁴ + 3x + 1, [−2, −1]