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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

Use formal definitions to prove the limit statements in Exercises 93–96.


lim x → 0 (−1 / x²) = −∞

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Textbook Question

Domains and Asymptotes


Determine the domain of each function in Exercises 69–72. Then use various limits to find the asymptotes.


y = (√(x² + 4)) / x

288
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Textbook Question

Infinite Limits


Find the limits in Exercises 37–48. Write ∞ or −∞ where appropriate.


lim x→0 (−1) / (x² (x + 1))

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Textbook Question

Slope of a Curve at a Point


In Exercises 7–18, use the method in Example 3 to find (a) the slope of the curve at the given point P, and (b) an equation of the tangent line at P.


y=x³−3x²+4, P(2,0)

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Textbook Question

Limits of Rational Functions


In Exercises 13–22, find the limit of each rational function (a) as x → ∞ and (b) as x → −∞. Write ∞ or −∞ where appropriate.


f(x) = (2x + 3)/(5x + 7)

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Textbook Question

Removable discontinuity Give an example of a function f (x) that is continuous for all values of x except x = 2, where it has a removable discontinuity. Explain how you know that f is discontinuous at x = 2, and how you know the discontinuity is removable.

499
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