The graph of a function f is shown. The x y-coordinate plane is given. The curve begins at y = 1 on the positive y-axis, appears to go horizontally right, passes through the approximate point (3, 1), goes up and right becoming more steep, and ends at the point (5, 3) nearly vertical. Does f satisfy the hypotheses of the Mean Value Theorem on the interval [0, 5]? Yes, because f is continuous on the open interval (0, 5) and differentiable on the closed interval [0, 5].Yes, because f is increasing on closed interval [0, 5]. Yes, because f is continuous on the closed interval [0, 5] and differentiable on the open interval (0, 5).No, because f is not differentiable on the open interval (0, 5).No, because f does not have a minimum nor a maximum on the closed interval [0, 5].No, because f is not continuous on the open interval (0, 5). If so, find a value c that satisfies the conclusion of the Mean Value Theorem on that interval. (If an answer does not exist, enter DNE.) c =

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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The graph of a function f is shown.
The x y-coordinate plane is given. The curve begins at y = 1 on the positive y-axis, appears to go horizontally right, passes through the approximate point (3, 1), goes up and right becoming more steep, and ends at the point (5, 3) nearly vertical.
Does f satisfy the hypotheses of the Mean Value Theorem on the interval [0, 5]?
Yes, because f is continuous on the open interval (0, 5) and differentiable on the closed interval [0, 5].Yes, because f is increasing on closed interval [0, 5].    Yes, because f is continuous on the closed interval [0, 5] and differentiable on the open interval (0, 5).No, because f is not differentiable on the open interval (0, 5).No, because f does not have a minimum nor a maximum on the closed interval [0, 5].No, because f is not continuous on the open interval (0, 5).
If so, find a value c that satisfies the conclusion of the Mean Value Theorem on that interval. (If an answer does not exist, enter DNE.)
c = 
3
 
 
 
0.
1
Transcribed Image Text:0. 1
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