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Ch. 3 - Polynomial and Rational Functions
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 4, Problema 35

Use the Intermediate Value Theorem to show that each polynomial has a real zero between the given integers.f(x)=2x4−4x2+1; between -1 and 0

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Recall the Intermediate Value Theorem (IVT), which states that if a function \(f\) is continuous on a closed interval \([a, b]\) and \(f(a)\) and \(f(b)\) have opposite signs, then there exists at least one \(c\) in \((a, b)\) such that \(f(c) = 0\).
Identify the function and the interval: here, \(f(x) = 2x^4 - 4x^2 + 1\) and the interval is \([-1, 0]\).
Evaluate the function at the endpoints of the interval: calculate \(f(-1)\) and \(f(0)\) by substituting these values into the polynomial.
Check the signs of \(f(-1)\) and \(f(0)\). If one is positive and the other is negative, then by the IVT, there is at least one real zero between \(-1\) and \(0\).
Conclude that since \(f\) is a polynomial (and thus continuous everywhere) and the function values at the endpoints have opposite signs, there must be a real zero of \(f(x)\) between \(-1\) and \(0\).

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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 values f(a) and f(b) at each end, then it must take any value between f(a) and f(b) at some point within the interval. This theorem is used to prove the existence of roots when the function changes sign.
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Polynomial Continuity

Polynomials are continuous functions for all real numbers, meaning there are no breaks, jumps, or holes in their graphs. This continuity ensures that the Intermediate Value Theorem can be applied to polynomials on any interval, making it possible to locate zeros by checking sign changes.
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Evaluating Function Values at Interval Endpoints

To apply the Intermediate Value Theorem, you must calculate the function's values at the given interval endpoints. If the function values at these points have opposite signs, it guarantees at least one root exists between them, as the function must cross the x-axis.
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