# SOLVED: Hw16-3.2-Mean-Value-Theorem: Problem 6 Problem Value: 6 points. Problem Score: 0%. Attempts Remaining: 12 attempts. Help Entering Answers point) Similar Example PDF The goal of this problem is

Mean Value Theorem For Integrals
Mean Value Theorem For Integrals

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Hw16-3.2-Mean-Value-Theorem: Problem 6
Problem Value: 6 points. Problem Score: 0%. Attempts Remaining: 12 attempts.
Similar Example PDF
The goal of this problem is to show that the function
f(x) = âˆš5
satisfies both of the conditions (the hypotheses and the conclusion) of the Mean Value Theorem for the interval [5,5].
Verification of Hypotheses: Fill in the blanks to show that the hypotheses of the Mean Value Theorem are satisfied.
f(x) is continuous on [5, 5] and differentiable on (-5, 5)
Note: The answer to each box should be one word.
Verification of the Conclusion: If the hypotheses of the Mean Value Theorem are satisfied, then there is at least one c in the interval (5, 5) for which
f(5) = f'(c)
Verify that the conclusion of the Mean Value Theorem holds by computing
f(5) – f(-5) / (-5)
Now find c in (-5, 5) so that f'(c) equals the answer you just found (For this problem, there is only one correct value of c)
Note: On an exam, YOU may be asked to state the Mean Value Theorem (i.e., it may not be given to you) or verify that a given function satisfies the assumptions of the Mean Value Theorem.

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Hw16-3.2-Mean-Value-Theorem: Problem 3Problem Value: 30 points. Problem Score: 030. Attempts Remaining: 12 attempts.Help Entering Answers point)Similar Example PDFThe goal of this problem is to show that the functionf(x) = âˆšxsatisfies both of the conditions (the hypotheses) and the conclusion of the Mean Value Theorem for the interval [-2, 2].

Verification Hypotheses: Fill in the blanks to show that the hypotheses of the Mean Value Theorem are satisfied f(s) = âˆšson [-2, 2] andf(s) = âˆšson (-2, 2).Note: The answer in each box should be one word.

Verification of the Conclusion: If the hypotheses of the Mean Value Theorem are satisfied, then there is at least one c in the interval (-2, 2) for which f'(c) = (f(2) – f(-2))/(2 – (-2)).

Verify that the conclusion of the Mean Value Theorem holds by computingf(2) – f(-2) = âˆš2 – âˆš(-2)

Now find c in (-2, 2) so that f'(c) equals the answer you just found.Note: For this problem, there are two correct values.We call them …

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f(x) =*satisfies both of the conditions (the hypotheses and the conclusion of the Mean Value Theorem for in the interval[5.9] Verification of Hypotheses: Fill in the blanks to show that the hypotheses of the Mean Value Theorem are satisfied: f(x) continuous 5.9] and differentiable (3.9) Note: The answer in each box should be one word:Verification of the Conclusion: If the hypotheses 0f the Mean Value Theorem are satisfied _ then there af least one in the interval(&,9)for whichVerify that the conclusion of the Mean Value Theorem holds by computingf(9)Now find9,9 so that f’ (c) equals the answe you just found: (For this problem there is only one correct value of c )Note: On a exam you may be asked to state the Mean Value Theorem (i.e it may not be given tO you); and to verify that given function satisfies the assumptions of the Mean Value Theorem:

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Given Mean-Value Theorem: Let f be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Then there is at least one point c in the interval (a,b) such that f'(c) = (f(b) – f(a))/(b – a).

Determine whether the Mean-Value Theorem can be applied to f on the closed interval [a,b]. If the Mean-Value Theorem can be applied, find all values of c in the open interval (a,b) such that f'(c) = (f(b) – f(a))/(b – a). If the Mean-Value Theorem cannot be applied, explain why not completely.

a) f(x) = -âˆš(x^2 + 8)b) f(x) = âˆš(5 – x^2); [-5,3]

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