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Chapter 7: Problem 68

If \(\mathrm{f}:[-1,1] \rightarrow \mathrm{R}\) is a continuously differetiable function such that \(f(1)>f(-1)\) and \(|P(y)| \leq 1\) for all \(y \in[-1,1]\) then (A) there exists an \(\mathrm{x} \in[-1,1]\) such that \(\mathrm{f}(\mathrm{x})>0\) (B) there exists an \(x \in[-1,1]\) such that \(f(x)<0\) (C) \(f(1) \leq f(-1)+2\) (D) \(f(-1), f(1)<0\)

Short Answer

Expert verified
(A) there exists an x in [-1, 1] such that f(x) > 0 (B) there exists an x in [-1, 1] such that f(x) < 0 (C) f(1) 鈮� f(-1) + 2 (D) f(-1), f(1) < 0 Answer: (C) f(1) 鈮� f(-1) + 2

Step by step solution

01

Understanding the given information

We are given a continuously differentiable function f on the interval [-1, 1]. This means that the function f is continuous and has a derivative in this interval. We also have the inequality f(1) > f(-1) and the constraint |P(y)| 鈮� 1 for all y in [-1, 1].
02

Using Mean Value Theorem (MVT)

Since the function f is continuous and has a derivative in [-1, 1], we can use the Mean Value Theorem (MVT). The MVT states that if a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in the open interval (a, b) such that f'(c) is equal to the average rate of change of the function in the interval: f'(c) = (f(b) - f(a)) / (b - a) In this case, let's apply MVT to f in the interval [-1, 1]: f'(c) = (f(1) - f(-1)) / (1 - (-1)) By substitution, we get: f'(c) = (f(1) - f(-1)) / 2
03

Evaluating the options

Now, let's analyze each of the given options: (A) there exists an x in [-1, 1] such that f(x) > 0 We cannot confirm this statement because to do so, we would have to know more about the function f. (B) there exists an x in [-1, 1] such that f(x) < 0 Again, we cannot confirm this statement because we would have to know more about the function f. (C) f(1) 鈮� f(-1) + 2 From the MVT, we know f'(c) = (f(1) - f(-1)) / 2. Since |P(y)| 鈮� 1 for all y in [-1, 1], and f'(c) is equal to P(c) for some c in [-1, 1], it follows that |f'(c)| 鈮� 1. This implies that: -f(1) + f(-1) 鈮� 2 and f(1) - f(-1) 鈮� 2 By adding f(-1) to both sides of the second inequality, we get: f(1) 鈮� f(-1) + 2 Hence, option (C) is correct. (D) f(-1), f(1) < 0 Since we have f(1) > f(-1), it is not possible for both f(-1) and f(1) to be less than 0. Therefore, option (D) is incorrect. So, based on our analysis, the correct choice is (C) f(1) 鈮� f(-1) + 2.

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Most popular questions from this chapter

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