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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 2.5.69
Chapter 2, Problem 2.5.69

Use the Intermediate Value Theorem in Exercises 69–74 to prove that each equation has a solution. Then use a graphing calculator or computer grapher to solve the equations.
x³ − 3x − 1 = 0

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1
Identify the function f(x) = x³ − 3x − 1 and note that it is a polynomial, which is continuous everywhere.
Choose an interval [a, b] where the function changes sign, indicating a root exists. For example, test f(0) and f(2).
Calculate f(0) = 0³ − 3(0) − 1 = -1 and f(2) = 2³ − 3(2) − 1 = 3. Since f(0) < 0 and f(2) > 0, there is a sign change.
Apply the Intermediate Value Theorem: Since f(x) is continuous on [0, 2] and f(0) < 0 < f(2), there exists at least one c in (0, 2) such that f(c) = 0.
Use a graphing calculator or computer grapher to approximate the root of the equation x³ − 3x − 1 = 0 within the interval [0, 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 continuous function f(x) takes on values f(a) and f(b) at two points a and b, and if f(a) and f(b) have opposite signs, then there exists at least one c in the interval (a, b) such that f(c) = 0. This theorem is crucial for proving the existence of a solution to the equation x³ − 3x − 1 = 0.
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Continuous Functions

A function is continuous if there are no breaks, jumps, or holes in its graph over its domain. For the Intermediate Value Theorem to apply, the function x³ − 3x − 1 must be continuous over the interval being considered. Polynomial functions, like this one, are continuous everywhere, which allows us to use the theorem effectively.
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Graphical Analysis

Graphical analysis involves using a graphing calculator or computer software to visualize the function and identify where it crosses the x-axis, indicating a root. By graphing x³ − 3x − 1, we can visually confirm the existence of a solution and approximate its value, complementing the theoretical proof provided by the Intermediate Value Theorem.
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Related Practice
Textbook Question

At what points are the functions in Exercises 13–30 continuous?

y = (2x – 1)¹/³

Textbook Question

At what points are the functions in Exercises 13–30 continuous?

y = √(x⁴ +1)/(1 + sin² x)

Textbook Question

Using the Formal Definition


Prove the limit statements in Exercises 37–50.


limx→−3 (x² − 9) / (x + 3) = −6

Textbook Question

In Exercises 1–4, say whether the function graphed is continuous on [−1, 3]. If not, where does it fail to be continuous and why?

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Textbook Question

In Exercises 77–80, find a function that satisfies the given conditions and sketch its graph. (The answers here are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.)


lim x → ±∞ k(x) = 1, lim x → 1⁻ k(x) = ∞, and lim x → 1⁺ k(x) = −∞

Textbook Question

Suppose that a function f(x) is defined for all x in [-1,1]. Can anything be said about the existence of limx→0 f(x)? Give reasons for your answer.