For the following excercise, show there is no c such that ?(1)−?(−1)=?′(?)(2).f(1)−f(−1)=f′(c)(2). Explain why the Mean Value Theorem does not apply over the interval [−1,1].[−1,1].F(x)= 1/x^2 12:39A openstax.org4.4 The Mean Value TheoremFor the following exercises, use the Mean Value Theoremand find all points 0 < c < 2 such thatf (2) – f (0) = f' (c) (2 – 0).161. f(x) = x³162. f(x) = sin(xx)163. f(x) = cos(2rx)164. f(x) = 1 +x + x?165. f(x) = (x – 1)'0166 f(x) =.(x- 1For the following exercises, show there is no c such thatf (1) – f (-1) = f' (c) (2). Explain why the Mean ValueTheorem does not apply over the interval [-1,1].167. 1 (X) = x --168. f(x) = -x2169. f(x) = Vlx|170. f (x) = [x] (Hint: This is called the floor functionand it is defined so that f(x) is the largest integer lessthan or equal to x.)For the following exercises, determine whether the MeanValue Theorem applies for the functions over the giveninterval [a, b]. Justify your answer.171. y = e* over [0, 1]172. y = In(2x + 3) over[-3,이173. f(x) = tan(2xx) over [0, 2]174. y = V6.v2 nver L_3 31.

Answered: Image /qna-images/answer/bf7209cf-bc51-4781-8a5b-4984b6fda165.jpg

Answered: f (1) – f (-1) = f' (c) (2). Explain…

Math

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f (1) – f (-1) = f' (c) (2). Explain why the Mean Value Theorem does not apply over the interval [-1, 1].

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter7: Analytic Trigonometry

Section7.6: The Inverse Trigonometric Functions

Problem 93E

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Concept explainers

Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topics, such as continuity and differentiability better. In a broader sense, every calculus notion is a limit. Other branches of calculus have also been derived from limits.

Question

For the following excercise, show there is no c such that ?(1)−?(−1)=?′(?)(2).f(1)−f(−1)=f′(c)(2). Explain why the Mean Value Theorem does not apply over the interval [−1,1].[−1,1].

F(x)= 1/x^2

12:39
A openstax.org
4.4 The Mean Value Theorem
For the following exercises, use the Mean Value Theorem
and find all points 0 < c < 2 such that
f (2) – f (0) = f' (c) (2 – 0).
161. f(x) = x³
162. f(x) = sin(xx)
163. f(x) = cos(2rx)
164. f(x) = 1 +x + x?
165. f(x) = (x – 1)'0
166 f(x) =.(x- 1
For the following exercises, show there is no c such that
f (1) – f (-1) = f' (c) (2). Explain why the Mean Value
Theorem does not apply over the interval [-1,1].
167. 1 (X) = x --
168. f(x) = -
x2
169. f(x) = Vlx|
170. f (x) = [x] (Hint: This is called the floor function
and it is defined so that f(x) is the largest integer less
than or equal to x.)
For the following exercises, determine whether the Mean
Value Theorem applies for the functions over the given
interval [a, b]. Justify your answer.
171. y = e* over [0, 1]
172. y = In(2x + 3) over
[-3,이
173. f(x) = tan(2xx) over [0, 2]
174. y = V
6.
v2 nver L_3 31.

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