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anonymous
 5 years ago
Consider x^2 on an interval [0,1/2]. The Mean Value theorem suggest that there is a number c in (0.1/1/2) such that f'(c) is equal to a particular d.What is d?
anonymous
 5 years ago
Consider x^2 on an interval [0,1/2]. The Mean Value theorem suggest that there is a number c in (0.1/1/2) such that f'(c) is equal to a particular d.What is d?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0find the slope of the secant line that passes through (.1, .01) and (.5, .25)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, my browser crashed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is the interval [0, 1/2] or [0.1, 1/2]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, I'll assume [0, 1/2] then. Let's say f(x) = x^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then the slope of the line between (0, f(0) and (.5, f(.5)) is (.250)/(.50) = 0.5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0next find where the derivative equals 0.5. This is the point where the tangent line is parallel to the secant line between (0, f(0) and (.5, f(.5))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry I crashed too haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so f'(x) = 2x = 0.5 x = 0.25

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh k, that makes more sense

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because f'(0.25) = 0.5 so c = 0.25 d = 0.5
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