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Ch. 2 - Limits and Continuity
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 2 - Limits and ContinuityProblem 2.29a
Chapter 2, Problem 2.29a

[Technology Exercise] Roots


Let Ζ’(𝓍) = 𝓍³ ―𝓍― 1.


a. Use the Intermediate Value Theorem to show that Ζ’ has a zero between ―1 and 2 .

Verified step by step guidance
1
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)Β³ - (-1) - 1. Simplify this expression to find the value of f(-1).
Calculate f(2): Substitute 𝓍 = 2 into the function: f(2) = (2)Β³ - (2) - 1. Simplify this expression to find the value of f(2).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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.
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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 Ζ’(𝓍) = 𝓍³ - 𝓍 - 1, are continuous everywhere, making them suitable for this theorem.
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Intro 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 Ζ’(-1) and Ζ’(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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Average Value of a Function
Related Practice
Textbook Question

Find the limits in Exercises 59–62. Write ∞ or βˆ’βˆž where appropriate.


lim ( 1 / xΒΉ/Β³ βˆ’ 1 / (x βˆ’ 1)⁴/Β³ ) as


a. x β†’ 0⁺

Textbook Question

Limits and Continuity

On what intervals are the following functions continuous?


a. Ζ’(x) = xΒΉ/Β³

Textbook Question

Exercises 5–10 refer to the function

f(x) = { xΒ² βˆ’ 1, βˆ’1 ≀ x < 0

2x, 0 < x < 1

1, x = 1

βˆ’2x + 4, 1 < x < 2

0, 2 < x < 3

graphed in the accompanying figure.

<IMAGE>

a. Does f (1) exist?

Textbook Question

Estimating Limits


[Technology Exercise] You will find a graphing calculator useful for Exercises 67–74.


Let f(x) = (xΒ² - 9) / (x + 3)


b. Support your conclusions in part (a) by graphing f near c = -3 and using Zoom and Trace to estimate y-values on the graph as x β†’ βˆ’3.

Textbook Question

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)=xΒ²βˆ’2x


a. [1, 3]

Textbook Question

Suppose that limxβ†’βˆ’2 p(x) = 4, limxβ†’βˆ’2 r(x) = 0, and limxβ†’βˆ’2 s(x) = βˆ’3. Find


a. limxβ†’βˆ’2 (p(x) + r(x) + s(x))