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Use intermediate theorem to explain P [x] = x* – x' – 10 has a zero in [2, 3]

Use intermediate theorem to explain P [x] = x* – x' – 10 has a zero in [2, 3]

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Title: Applying the Intermediate Value Theorem** **Content:** **Concept: Intermediate Value Theorem (IVT)** The Intermediate Value Theorem states that for any continuous function \( f(x) \) over a closed interval \([a, b]\), if \( f(a) \) and \( f(b) \) have opposite signs, then there is at least one \( c \) in \((a, b)\) such that \( f(c) = 0 \). **Problem:** Use the Intermediate Value Theorem to explain why the polynomial function \( P(x) = x^4 - x^3 - 10 \) has a zero in the interval \([2, 3]\). **Solution:** 1. **Check the Continuity:** - The function \( P(x) = x^4 - x^3 - 10 \) is a polynomial, and polynomials are continuous everywhere. Therefore, it is continuous on the closed interval \([2, 3]\). 2. **Evaluate the Function at the Endpoints:** - Calculate \( P(2) \): \[ P(2) = 2^4 - 2^3 - 10 = 16 - 8 - 10 = -2 \] - Calculate \( P(3) \): \[ P(3) = 3^4 - 3^3 - 10 = 81 - 27 - 10 = 44 \] 3. **Check for a Sign Change:** - \( P(2) = -2 \) and \( P(3) = 44 \). The signs are different (\(-2\) is negative and \(44\) is positive). 4. **Apply the Intermediate Value Theorem:** - Since \( P(x) \) is continuous on \([2, 3]\) and \( P(2) \) and \( P(3) \) have opposite signs, the IVT guarantees that there is at least one value \( c \) in the interval \((2, 3)\) where \( P(c) = 0 \). **Conclusion:** By using the Intermediate Value Theorem, we have demonstrated that the polynomial function \( P(x) = x^4 - x^3 - 10 \) has at least one zero in the interval \
Transcribed Image Text:**Title: Applying the Intermediate Value Theorem** **Content:** **Concept: Intermediate Value Theorem (IVT)** The Intermediate Value Theorem states that for any continuous function \( f(x) \) over a closed interval \([a, b]\), if \( f(a) \) and \( f(b) \) have opposite signs, then there is at least one \( c \) in \((a, b)\) such that \( f(c) = 0 \). **Problem:** Use the Intermediate Value Theorem to explain why the polynomial function \( P(x) = x^4 - x^3 - 10 \) has a zero in the interval \([2, 3]\). **Solution:** 1. **Check the Continuity:** - The function \( P(x) = x^4 - x^3 - 10 \) is a polynomial, and polynomials are continuous everywhere. Therefore, it is continuous on the closed interval \([2, 3]\). 2. **Evaluate the Function at the Endpoints:** - Calculate \( P(2) \): \[ P(2) = 2^4 - 2^3 - 10 = 16 - 8 - 10 = -2 \] - Calculate \( P(3) \): \[ P(3) = 3^4 - 3^3 - 10 = 81 - 27 - 10 = 44 \] 3. **Check for a Sign Change:** - \( P(2) = -2 \) and \( P(3) = 44 \). The signs are different (\(-2\) is negative and \(44\) is positive). 4. **Apply the Intermediate Value Theorem:** - Since \( P(x) \) is continuous on \([2, 3]\) and \( P(2) \) and \( P(3) \) have opposite signs, the IVT guarantees that there is at least one value \( c \) in the interval \((2, 3)\) where \( P(c) = 0 \). **Conclusion:** By using the Intermediate Value Theorem, we have demonstrated that the polynomial function \( P(x) = x^4 - x^3 - 10 \) has at least one zero in the interval \
Expert Solution
Step 1

According to intermediate value theorem of a function f(x) is continuous on [a, b] and L is any number between f(a) and f(b) then there must be a value x= c where a<c<b such that f(c) =L.

Here the given function is P(x) = x4—x3—10

Since the given function is a polynomial function, we know that all polynomial functions are continuous everywhere so the given polynomial function is also continuous on [2,3]. So here we can use intermediate value theorem.

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