anonymous
  • anonymous
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.
Mathematics
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katieb
  • katieb
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anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
That's Rolle's theorem.

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anonymous
  • anonymous
oh m wrong :(
anonymous
  • anonymous
:)
anonymous
  • anonymous
yes lokisan u r right
anonymous
  • anonymous
Is that all, Mr President?
anonymous
  • anonymous
thank you very much lokisan. you are a lifesaver! :)
anonymous
  • anonymous
You're welcome :)
anonymous
  • anonymous
one more question though lokisan... how did you come up wit the derivative being -1/x^2?
anonymous
  • anonymous
\[f(x)=\frac{x+1}{x}=1+\frac{1}{x} \rightarrow f'(x)=\frac{d}{dx}0+\frac{d}{dx}x^{-1}=0-x^{-2}=-\frac{1}{x^2}\]
anonymous
  • anonymous
It should read \[\frac{d}{dx}1\]not\[\frac{d}{dx}0\]
anonymous
  • anonymous
omg why didnt i think of that. i was trying to do the quotient rule this whole time!
anonymous
  • anonymous
Yeah, try and simplify if you can.

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