Consider The Function F(x)=1/x On The Interval [3,

CSRoger2

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Consider the function f(x)=1/x on the interval [3,11]. Find the average or mean slope of the function on this interval.???? By the Mean Value Theorem, we know there exists a c in the open interval (3,11) such that f(c) is equal to this mean slope. For this problem, there is only one c that works. Find it.????

 

Answer No.1

Anonymous

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Average slope =f'(c) = (f(11)-f(3))/(11-3) = -1/33 f'(c) = -1/c2 = -1/33 ∴c=√33 = 5.74

Answer No.2

Anonymous

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F'(x) = -1/x^2 mean slope = f(11)- f(3)/8 = -1/33 f'(c) = -1/c^2 = -1/33 c = 5.744

Answer No.3

Anonymous

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F1(C) = f(11)-f(3) / (11-3) = 1/11 - 1/3 /(8) = -8/(33x8) = -1/33 f1(c) = -1/c2 = -1/33 c2 = 33 --> c = 5.74 or √33

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