CSRoger2 From: US Posts: 27 Votes: 54 |
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.???? |

Anonymous From: - Posts: - Votes: - |
Average slope =f'(c) = (f(11)-f(3))/(11-3) = -1/33 f'(c) = -1/c2 = -1/33 ∴c=√33 = 5.74 |

Anonymous From: - Posts: - Votes: - |
F'(x) = -1/x^2 mean slope = f(11)- f(3)/8 = -1/33 f'(c) = -1/c^2 = -1/33 c = 5.744 |

Anonymous From: - Posts: - Votes: - |
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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