Understanding Mean Value Theorem: Solving Homework Problems

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Homework Statement


http://img14.imageshack.us/img14/6132/proiqc.jpg


Homework Equations





The Attempt at a Solution


the first 3 are from the textbook so they must be right.. the last 2 I am pretty sure i got right too..
because the 4th one, if f'(x)=0 then f(x)= c .. so its false.
im not too sure about the 5th one but I am 90% sure it should true.
 
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(1) is Rolle's Theorem (Edit: this is incorrect - see the following posts)
(2) is the Mean Value Theorem, which is a generalization of Rolle's Theorem
I did not know (3) but that is interesting
(4) is correct, f(x) can be a constant function not equal to 0.
(5) is correct - take a(x) = f(x) - g(x). It's derivative is 0, so a(x) is constant (you can prove this using MVT).
 
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VeeEight said:
(1) is Rolle's Theorem
(2) is the Mean Value Theorem, which is a generalization of Rolle's Theorem
I did not know (3) but that is interesting
(4) is correct, f(x) can be a constant function not equal to 0.
(5) is correct - take a(x) = f(x) - g(x). It's derivative is 0, so a(x) is constant (you can prove this using MVT).

i know that 1 and 2 are Rolle's Theorem and Mean Value Theorem, and its written exactly like in my textbook.. i can't find the mistake.

https://www.physicsforums.com/library.php?do=view_item&itemid=231 its even stated here..

so what is wrong?
 
Oh sorry, I missed that
(1) states that f is defined on [a,b] but it is not necessarily continuous there. It is continuous on (a,b) (since it is differentiable there) but not necessarily at a or b.
 
VeeEight said:
Oh sorry, I missed that
(1) states that f is defined on [a,b] but it is not necessarily continuous there. It is continuous on (a,b) (since it is differentiable there) but not necessarily at a or b.

thats a tricky question.. weird.
thanks a lot for your help man!
 
No problem, can't believe I missed that
In my experience, the physical science courses are where they try to 'trick' you like that, I've never had any questions like that when I was a math undergrad. I guess now you've learned that you have to pay attention to every little detail when writing tests in this class (which is probably a good idea, regardless). Cheers.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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