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Chapter 10: Problem 865

The Roll's theorem is applicable in the interval \(-1 \leq \mathrm{x} \leq 1\) for the function (a) \(\mathrm{f}(\mathrm{x})=\mathrm{x}\) (b) \(f(x)=x^{2}\) (c) \(f(x)=2 x^{2}+3\) (d) \(\mathrm{f}(\mathrm{x})=|\mathrm{x}|\)

Short Answer

Expert verified
Rolle's theorem is applicable for \(f(x) = x^2\) and \(f(x) = 2x^2 + 3\) in the interval \([-1,1]\), as they satisfy all the conditions of Rolle's theorem. However, it is not applicable for \(f(x) = x\) and \(f(x) = |x|\), as they do not satisfy all the required conditions.

Step by step solution

01

(a) Check conditions for f(x) = x

First, let's consider the function f(x) = x. 1. Continuity: Since f(x) = x is a polynomial function, it is continuous in the entire real line, including the interval [-1, 1]. 2. Differentiability: Similarly, since f(x) = x is a polynomial function, it is differentiable in the entire real line, including the interval (-1, 1). 3. Endpoints: f(-1) = -1 and f(1) = 1. Since f(-1) ≠ f(1), the function f(x) = x does not satisfy the third condition of Rolle's theorem. Therefore, we cannot apply Rolle's theorem to function f(x) = x in the interval [-1, 1].
02

(b) Check conditions for f(x) = x^2

Now, let's consider the function f(x) = x^2. 1. Continuity: Since f(x) = x^2 is a polynomial function, it is continuous in the entire real line, including the interval [-1, 1]. 2. Differentiability: Similarly, since f(x) = x^2 is a polynomial function, it is differentiable in the entire real line, including the interval (-1, 1). 3. Endpoints: f(-1) = 1 and f(1) = 1. Since f(-1) = f(1), the function f(x) = x^2 satisfies all the conditions of Rolle's theorem, and we can apply Rolle's theorem to function f(x) = x^2 in the interval [-1, 1].
03

(c) Check conditions for f(x) = 2x^2 + 3

Let's consider the function f(x) = 2x^2 + 3. 1. Continuity: Since f(x) = 2x^2 + 3 is a polynomial function, it is continuous in the entire real line, including the interval [-1, 1]. 2. Differentiability: Similarly, since f(x) = 2x^2 + 3 is a polynomial function, it is differentiable in the entire real line, including the interval (-1, 1). 3. Endpoints: f(-1) = 2(-1)^2 + 3 = 5 and f(1) = 2(1)^2 + 3 = 5. Since f(-1) = f(1), the function f(x) = 2x^2 + 3 satisfies all the conditions of Rolle's theorem, and we can apply Rolle's theorem to function f(x) = 2x^2 + 3 in the interval [-1, 1].
04

(d) Check conditions for f(x) = |x|

Finally, let's consider the function f(x) = |x|. 1. Continuity: Since f(x) = |x| is a continuous function in the entire real line, it is continuous in the interval [-1, 1]. 2. Differentiability: f(x) = |x| is not differentiable at x = 0, since its derivative is not defined in that point (the function has a sharp turn at x = 0). Therefore, f(x) = |x| does not satisfy the second condition of Rolle's theorem. 3. Endpoints: f(-1) = |-1| = 1 and f(1) = |1| = 1. The function f(x) = |x| satisfies the third condition of Rolle's theorem. However, since f(x) = |x| does not satisfy the second condition (differentiability), we cannot apply Rolle's theorem to function f(x) = |x| in the interval [-1, 1].

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