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Chapter 3: Problem 84

If \(f\) is a polynomial function, and \(f(a)\) and \(f(b)\) have opposite signs, what must occur between \(a\) and \(b ?\) If \(f(a)\) and \(f(b)\) have the same sign, does it necessarily mean that this will not occur? Explain your answer.

Short Answer

Expert verified
In the case of \(f(a)\) and \(f(b)\) having opposing signs, there must be a root, or zero, of the function between \(a\) and \(b\). On the other hand, having the same sign at \(a\) and \(b\) does not rule out the possibility of a root between the two points.

Step by step solution

01

Analyze the first scenario

The first part of the exercise implies that \(f(a)\) and \(f(b)\) have opposing signs. According to the Intermediate Value Theorem (IVT), if \(f\) is a polynomial (hence a continuous function), and \(f(a)\) and \(f(b)\) have differing signs, then \(f(x)\) must be equal to 0 for at least one \(x\) in the interval between \(a\) and \(b\). Therefore, if \(f(a)\) and \(f(b)\) are opposing in sign, there must be a root, or zero, of the function between \(a\) and \(b\).
02

Analyze the second scenario

The second part of the question asks whether the same condition will not occur if \(f(a)\) and \(f(b)\) have the same sign. The answer is: not necessarily. If \(f(a)\) and \(f(b)\) are equal in sign, it does not imply there is no root between \(a\) and \(b\). For instance, the function could cross the x-axis and then return to the same side, resulting in roots even though the signs at points \(a\) and \(b\) are the same. Thus, while the Intermediate Value Theorem (IVT) can guarantee a root when \(f(a)\) and \(f(b)\) have opposing signs, it cannot discount the possibility of roots when they have the same sign.

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