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anonymous
 5 years ago
Determine if the function f(x)=(x+1)/x satisfies the conditions of the Mean Value Theorem on [1,4], and find all value(s) of c in (1,4) such that f'(c)=(f(b)f(a))/ba.
anonymous
 5 years ago
Determine if the function f(x)=(x+1)/x satisfies the conditions of the Mean Value Theorem on [1,4], and find all value(s) of c in (1,4) such that f'(c)=(f(b)f(a))/ba.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The Mean Value Theorem says that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that\[f'(c)=\frac{f(b)f(a)}{ba}\]Your function f is continuous and differentiable on\[[1,4] ,(1,4)\]respectively. You need to now find all c that satisfy the theorem:\[f'(x)=\frac{1}{x^2} \rightarrow f'(c)=\frac{1}{c^2}\]that is,\[\frac{1}{c^2}=\frac{5/42}{41} \rightarrow c^2=4 \rightarrow c = \pm 2\]Since\[2 \notin ([1,4]\]you take c=2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the given function does not satisfy the conditions, function is continuous on [1,4], differntiable on (1,4) but f(a) = f(b) is not satisfied

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's Rolle's theorem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes lokisan u r right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is that all, Mr President?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you very much lokisan. you are a lifesaver! :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one more question though lokisan... how did you come up wit the derivative being 1/x^2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=\frac{x+1}{x}=1+\frac{1}{x} \rightarrow f'(x)=\frac{d}{dx}0+\frac{d}{dx}x^{1}=0x^{2}=\frac{1}{x^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It should read \[\frac{d}{dx}1\]not\[\frac{d}{dx}0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0omg why didnt i think of that. i was trying to do the quotient rule this whole time!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, try and simplify if you can.
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