a. Determine whether the Mean Value Theorem applies to the function f(x) = x + on the interval [1,6]. Xx b. If so, find or approximate the point(s) that are guaranteed to exist by the Mean Value Theorem. a. Choose the correct answer below. A. No, because the function is not continuous on the interval [1,6], and is not differentiable on the interval (1,6). B. No, because the function is continuous on the interval [1,6], but is not differentiable on the interval (1,6). C. Yes, because the function is continuous on the interval [1,6] and differentiable on the interval (1,6). D. No, because the function is differentiable on the interval (1,6), but is not continuous on the interval [1,6]. b.Select the correct choice below and, if necessary, fill in the answer box to complete your choice. OA. *** 1800
a. Determine whether the Mean Value Theorem applies to the function f(x) = x + on the interval [1,6]. Xx b. If so, find or approximate the point(s) that are guaranteed to exist by the Mean Value Theorem. a. Choose the correct answer below. A. No, because the function is not continuous on the interval [1,6], and is not differentiable on the interval (1,6). B. No, because the function is continuous on the interval [1,6], but is not differentiable on the interval (1,6). C. Yes, because the function is continuous on the interval [1,6] and differentiable on the interval (1,6). D. No, because the function is differentiable on the interval (1,6), but is not continuous on the interval [1,6]. b.Select the correct choice below and, if necessary, fill in the answer box to complete your choice. OA. *** 1800
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Please help with part b!
![1
a. Determine whether the Mean Value Theorem applies to the function f(x)=x+ on the interval [1,6].
X
b. If so, find or approximate the point(s) that are guaranteed to exist by the Mean Value Theorem.
a. Choose the correct answer below.
OA. No, because the function is not continuous on the interval [1,6], and is not differentiable on the interval (1,6).
B. No, because the function is continuous on the interval [1,6], but is not differentiable on the interval (1,6).
C. Yes, because the function is continuous on the interval [1,6] and differentiable on the interval (1,6).
D. No, because the function is differentiable on the interval (1,6), but is not continuous on the interval [1,6].
b. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
4899
The point(s) is/are x =
2000
(Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.)
B. The Mean Value Theorem does not apply in this case.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa204fb01-1593-427c-b210-f132b42420bd%2F44ea9acb-2664-49bf-9df9-d3cbacbeddf4%2F3ls8cym_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1
a. Determine whether the Mean Value Theorem applies to the function f(x)=x+ on the interval [1,6].
X
b. If so, find or approximate the point(s) that are guaranteed to exist by the Mean Value Theorem.
a. Choose the correct answer below.
OA. No, because the function is not continuous on the interval [1,6], and is not differentiable on the interval (1,6).
B. No, because the function is continuous on the interval [1,6], but is not differentiable on the interval (1,6).
C. Yes, because the function is continuous on the interval [1,6] and differentiable on the interval (1,6).
D. No, because the function is differentiable on the interval (1,6), but is not continuous on the interval [1,6].
b. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
4899
The point(s) is/are x =
2000
(Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.)
B. The Mean Value Theorem does not apply in this case.
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