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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

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

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

That's Rolle's theorem.

4. anonymous

oh m wrong :(

5. anonymous

:)

6. anonymous

yes lokisan u r right

7. anonymous

Is that all, Mr President?

8. anonymous

thank you very much lokisan. you are a lifesaver! :)

9. anonymous

You're welcome :)

10. anonymous

one more question though lokisan... how did you come up wit the derivative being -1/x^2?

11. anonymous

$f(x)=\frac{x+1}{x}=1+\frac{1}{x} \rightarrow f'(x)=\frac{d}{dx}0+\frac{d}{dx}x^{-1}=0-x^{-2}=-\frac{1}{x^2}$

12. anonymous

It should read $\frac{d}{dx}1$not$\frac{d}{dx}0$

13. anonymous

omg why didnt i think of that. i was trying to do the quotient rule this whole time!

14. anonymous

Yeah, try and simplify if you can.

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