Domanda del libro di testo
Average Rates of Change
In Exercises 1–6, find the average rate of change of the function over the given interval or intervals.
h(t)=cot t
a. [π/4,3π/4]
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Ch. 2 - Limits and Continuity
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- 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 Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 2, Problema 2.29a
[Technology Exercise] Roots
Let ƒ(𝓍) = 𝓍³ ―𝓍― 1.
a. Use the Intermediate Value Theorem to show that ƒ has a zero between ―1 and 2 .
Guida verificata passo dopo passo1
First, understand the Intermediate Value Theorem (IVT): It states that if a function f is continuous on a closed interval [a, b] and N is any number between f(a) and f(b), then there exists at least one c in the interval (a, b) such that f(c) = N.
Check the continuity of the function f(𝓍) = 𝓍³ - 𝓍 - 1. Since this is a polynomial function, it is continuous everywhere, including on the interval [-1, 2].
Evaluate the function at the endpoints of the interval: Calculate f(-1) and f(2).
Calculate f(-1): Substitute 𝓍 = -1 into the function: f(-1) = (-1)³ - (-1) - 1. Simplify this expression to find the value of f(-1).
Calculate f(2): Substitute 𝓍 = 2 into the function: f(2) = (2)³ - (2) - 1. Simplify this expression to find the value of f(2).
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Intermediate Value Theorem
The Intermediate Value Theorem states that if a function is continuous on a closed interval [a, b] and takes on different signs at the endpoints, then there exists at least one c in (a, b) such that f(c) = 0. This theorem is crucial for proving the existence of roots within a specified interval.
Video consigliato:
Percorso guidato
06:11
Fundamental Theorem of Calculus Part 1
Continuity of Functions
A function is continuous if there are no breaks, jumps, or holes in its graph. For the Intermediate Value Theorem to apply, the function must be continuous over the interval in question. Polynomial functions, like ƒ(𝓍) = 𝓍³ - 𝓍 - 1, are continuous everywhere, making them suitable for this theorem.
Video consigliato:
05:34Intro to Continuity
Evaluating Function Values
To apply the Intermediate Value Theorem, we need to evaluate the function at the endpoints of the interval. By calculating ƒ(-1) and ƒ(2), we can determine the signs of these values. If they are of opposite signs, it confirms the existence of at least one root in the interval (-1, 2).
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Percorso guidato
06:37Average Value of a Function
Pratica correlata
Domanda del libro di testo
Limits and Continuity
On what intervals are the following functions continuous?
a. ƒ(x) = x¹/³
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Domanda del libro di testo
Find the limits in Exercises 59–62. Write ∞ or −∞ where appropriate.
lim ( 1 / x²/³ + 2 / (x − 1)²/³ ) as
a. x → 0⁺
274
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Domanda del libro di testo
Limits and Continuity
On what intervals are the following functions continuous?
a. ƒ(x) = tan x
266
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Domanda del libro di testo
Average Rates of Change
In Exercises 1–6, find the average rate of change of the function over the given interval or intervals.
g(x)=x²−2x
a. [1, 3]
429
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Domanda del libro di testo
Suppose that limx→−2 p(x) = 4, limx→−2 r(x) = 0, and limx→−2 s(x) = −3. Find
a. limx→−2 (p(x) + r(x) + s(x))
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