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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))/b-a.

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  1. anonymous
    • 5 years ago
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    The 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\[f'(c)=\frac{f(b)-f(a)}{b-a}\]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,\[-\frac{1}{c^2}=\frac{5/4-2}{4-1} \rightarrow c^2=4 \rightarrow c = \pm 2\]Since\[-2 \notin ([1,4]\]you take c=2.

  2. anonymous
    • 5 years ago
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    the 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

  3. anonymous
    • 5 years ago
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    That's Rolle's theorem.

  4. anonymous
    • 5 years ago
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    oh m wrong :(

  5. anonymous
    • 5 years ago
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    :)

  6. anonymous
    • 5 years ago
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    yes lokisan u r right

  7. anonymous
    • 5 years ago
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    Is that all, Mr President?

  8. anonymous
    • 5 years ago
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    thank you very much lokisan. you are a lifesaver! :)

  9. anonymous
    • 5 years ago
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    You're welcome :)

  10. anonymous
    • 5 years ago
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    one more question though lokisan... how did you come up wit the derivative being -1/x^2?

  11. anonymous
    • 5 years ago
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    \[f(x)=\frac{x+1}{x}=1+\frac{1}{x} \rightarrow f'(x)=\frac{d}{dx}0+\frac{d}{dx}x^{-1}=0-x^{-2}=-\frac{1}{x^2}\]

  12. anonymous
    • 5 years ago
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    It should read \[\frac{d}{dx}1\]not\[\frac{d}{dx}0\]

  13. anonymous
    • 5 years ago
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    omg why didnt i think of that. i was trying to do the quotient rule this whole time!

  14. anonymous
    • 5 years ago
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    Yeah, try and simplify if you can.

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