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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 4, Problema 4.2.22

Roots (Zeros)


Show that the functions in Exercises 19–26 have exactly one zero in the given interval.


g(t) = 1/(1 − t) + √(1 + t) − 3.1, (−1, 1)

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First, understand that finding the zero of a function means finding the value of t for which g(t) = 0. We need to show that g(t) has exactly one zero in the interval (-1, 1).
Consider the function g(t) = \(\frac{1}{1 - t}\) + \(\sqrt{1 + t}\) - 3.1. We need to analyze the behavior of this function within the interval (-1, 1).
Check the continuity of g(t) in the interval (-1, 1). The function is continuous in this interval because both \(\frac{1}{1 - t}\) and \(\sqrt{1 + t}\) are continuous for t in (-1, 1).
Use the Intermediate Value Theorem, which states that if a function is continuous on a closed interval [a, b] and takes different signs at the endpoints, then it has at least one zero in the interval (a, b). Evaluate g(t) at the endpoints of the interval: g(-1) and g(1).
Calculate the derivative g'(t) to determine the behavior of g(t) within the interval. If g'(t) does not change sign, it indicates that g(t) is either strictly increasing or strictly decreasing, which implies exactly one zero in the interval (-1, 1).

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Intermediate Value Theorem

The Intermediate Value Theorem states that if a continuous function, f, takes on values of opposite signs at two points, a and b, then it must have at least one root in the interval (a, b). This theorem is crucial for proving the existence of a zero in a given interval, as it ensures that the function crosses the x-axis.
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Continuity of Functions

A function is continuous on an interval if there are no breaks, jumps, or holes in its graph within that interval. For the function g(t) = 1/(1 − t) + √(1 + t) − 3.1, it is important to verify its continuity on the interval (−1, 1) to apply the Intermediate Value Theorem effectively. Discontinuities would invalidate the theorem's application.
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Behavior of Rational and Radical Functions

Understanding the behavior of rational and radical functions is essential for analyzing g(t). The function 1/(1 − t) is undefined at t = 1, and √(1 + t) is defined for t ≥ -1. Analyzing these components helps determine the function's behavior and continuity over the interval (−1, 1), ensuring it meets the conditions for the Intermediate Value Theorem.
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