Tough mean value theorem problem

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1) By applying the Mean Value Theorem to f(x)=sqrt x, show that 1/11 < (sqrt 102) -10 < 1/10.

This is a sample problem in my course, and how to start this problem is the key thing.
The solution says "f is continuous and differentiable for all x>0, so by the mean value theorem, there exists c E (100,102) such that (conclusion of mean value theorem)....etc"

Say if you have never seen such a problem before, how can you get the inspiration to pick the interval (100,102) and not something else? How is it even possible to know this ahead of time before you start the proof? Can someone teach me the logic of this choice and most importantly, how you arrived at the choice of picking this particular interval? Thank you very much!
 

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HallsofIvy
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1) By applying the Mean Value Theorem to f(x)=sqrt x, show that 1/11 < (sqrt 102) -10 < 1/10.

This is a sample problem in my course, and how to start this problem is the key thing.
The solution says "f is continuous and differentiable for all x>0, so by the mean value theorem, there exists c E (100,102) such that (conclusion of mean value theorem)....etc"

Say if you have never seen such a problem before, how can you get the inspiration to pick the interval (100,102) and not something else? How is it even possible to know this ahead of time before you start the proof? Can someone teach me the logic of this choice and most importantly, how you arrived at the choice of picking this particular interval? Thank you very much!
Your problem says [itex]1/11< \sqrt{102}- 10< 1/10[/itex]. How much "inspiration" does it take to see that [itex]10= \sqrt{100}[/itex]. Certainly, if I see something involving [itex]\sqrt{102}[/itex] and [itex]\sqrt{100}[/itex], it wouldn't take me long to think, "Hey, I'll try applying [itex]\sqrt{x}[/itex] to the interval from 100 to 102!"
 
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ok, I think you are right, it seems easier now...

Can I ask another question?

Are lim (x->c+) f'(x) and f'+(x)=lim (h->0+) [f(x+h)-f(x)]/h exactly the same thing? (namely, the right hand derivative of f)?

I have seen someone writing lim (x->c+) f'(x), but I have never seen someone writing in this form...so I don't quite understand its meaning, and whether or not it is equal to the right hand derivative of f??
 
HallsofIvy
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If f has a derivative at c, then, yes, they are the same thing. The derivative of a function is not necessarily continuous (in which case it would follow immediately that the two one sided limits are equal), but it does satisfy the "intermediate value property" (If f(a)= p and f(b)= q, then f takes on every value between p and q on the interval (a, b).). If the derivative exists, then the two one sided limits must exist and be the same.
 
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if the question was posted like this will it be enough to answer the question. Thanks in advanced

1. Use the Mean Value Theorem to prove that 1/11<sqrt(102)-10<1/10
 
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ok, I think you are right, it seems easier now...

Can I ask another question?

Are lim (x->c+) f'(x) and f'+(x)=lim (h->0+) [f(x+h)-f(x)]/h exactly the same thing? (namely, the right hand derivative of f)?

I have seen someone writing lim (x->c+) f'(x), but I have never seen someone writing in this form...so I don't quite understand its meaning, and whether or not it is equal to the right hand derivative of f??
No, they are not the same thing in all cases.

lim (x->c+) f'(x) is the limit of the derivative of f as x approaches c from the right.

f'+(x)=lim (h->0+) [f(x+h)-f(x)]/h is the right-hand derivative of f. I'm assuming you mean that x=c here.

They will be the same if f' is continuous, but derivatives are not necessarily continuous.

[Edit] For a counterexample, consider f(x) = x^2 sin(1/x) if x /= 0, f(0) = 0. Then f'(0) exists and is 0, but lim x->0 f'(x) does not exist.

As Halls said, they are the same IF both limits exist. But that's not necessarily the case.
[/edit]
 
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