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Chapter 4: Problem 17

Let \(f:[a, b] \rightarrow \mathbb{R}\) be continuous on \([a, b]\) and differentiable on \((a, b)\). If \(f(a)\) and \(f(b)\) are of different signs and \(f^{\prime}(x) \neq 0\) for all \(x \in(a, b)\), then show that there is a unique \(x_{0} \in(a, b)\) such that \(f\left(x_{0}\right)=0\).

Short Answer

Expert verified
By the Intermediate Value Theorem, there exists at least one \(x_{0} \in (a, b)\) such that \(f\left(x_{0}\right) = 0\). Assuming there exist two distinct points \(x_1, x_2 \in (a, b)\) such that \(f(x_1) = f(x_2) = 0\), Rolle's theorem implies the existence of \(c \in (x_1, x_2)\) such that \(f'(c) = 0\), contradicting the given condition. Hence, there exists a unique \(x_{0} \in (a, b)\) such that \(f\left(x_{0}\right) = 0\).

Step by step solution

01

Use the Intermediate Value Theorem (IVT) to find the existence of x_0

Given that f is continuous on [a, b], and f(a) and f(b) have different signs, by the Intermediate Value Theorem, there exists at least one \(x_{0} \in (a, b)\) such that \(f\left(x_{0}\right) = 0\).
02

Prove the uniqueness using the derivative conditions

In order to prove the uniqueness of x_0, let's assume there exist two distinct points \(x_1, x_2 \in (a, b)\) such that \(f(x_1) = f(x_2) = 0\). Then, applying Rolle's theorem on the interval [x_1, x_2] for the function f, since f is continuous, differentiable and f(x_1) = f(x_2) = 0, there must exist \(c \in (x_1, x_2)\) such that \(f'(c) = 0\). However, this contradicts the given condition \(f^{\prime}(x) \neq 0\) for all \(x \in(a, b)\). This contradiction implies that our assumption of two distinct points x_1 and x_2 with f(x_1) = f(x_2) = 0 is false. Therefore, there must be a unique \(x_0 \in (a, b)\) such that \(f\left(x_{0}\right) = 0\).

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Most popular questions from this chapter

Let \(c \in \mathbb{R}, r>0, f:(c-r, c+r) \rightarrow \mathbb{R}\), and \(n \in \mathbb{N}\) be such that \(f^{(n)}(c)\) exists. Show that \(\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)-h f^{\prime}(c)-\cdots-h^{n-1}\left[f^{(n-1)}(c) /(n-1) !\right]}{h^{n}}=\frac{f^{(n)}(c)}{n !}\)

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