Question

Mean Value Theorem Problem

Answer 1

Question

Answer 2

really? here is my proof, see if i have a mistake
on \((0,1)\) the function \(f(x)=\sin(x)>0\)
has all order continuous derivatives and \(f''(x)=-\sin(x)\) so \(|\frac{f''(x)}{f(x)}|=1\) and
\[\int_0^11dx=1<4\]

Answer 3

\[\sin 1\neq0\]

Answer 4

i agree but it says \(f(x)>0\) on \((0,1)\) doesn't say it is zero at the endpoints

Answer 5

i believe that i might be missing something, but i don't see what it is

Answer 6

ooooooooooooooooh nvm i was missing scrolling to the right to see the rest
sorry!

Answer 7

proof integral >4 or <4 ?

Answer 8

>4

Answer 9

i am lost on this one. if i think of something i will come back. honestly i cannot see why it should be true. i was providing a counter example

Answer 10

basically you are trying to prove that
\[|\frac{f''(x)}{f(x)}|>4\] which will bound the integral, but i don't see why this has to be true.
mvt for integrals says for some \[c\in (0,1), \frac{f''(c)}{f(c)}=\int_0^1\frac{f''(x)}{f(x)}dx\]

Answer 11

if you get an answer i would love to see it

Answer 12

likewise
*bookmark

Answer 13

whats proof?????

Answer 14

seems all right ... at-lest on this case
http://www.wolframalpha.com/input/?i=integrate+abs%282%2F%28x%28x-1%29%29%29+from+0.00001+to+0.99999

Answer 15

http://www.wolframalpha.com/input/?i=integrate+abs%282%2F%28500000x%28x-1%29%29%29+from+0+to+1

Answer 16

@Zarkon has been lingering on this problem for a long time...

Answer 17

I'm quite interested myself. Cannot think of how MVT applies here.

Answer 18

f(0)=f(1)=0
f(x) > 0 for all x∈(0,1)

Hidden Condition
f(x) ≤ f(x_o) for all x∈(0,1)

Answer 19

By MVT, ∃a∈(0,x_o) and ∃b∈(x_o,1) such that
f'(a)=f(x_o)/x_o
f'(b)=-f(x_o)/(1-x_o)

Answer 20

Can I proof like this?
\[\int\limits\limits\limits_{0}^{1}\left| f''(x)/f(x) \right|dx \ge \int\limits\limits\limits_{a}^{b}\left| f''(x)/f(x) \right|dx \ge \left| 1/f(x_o) \right| \left| \int\limits\limits\limits_{a}^{b}f''(x)dx \right|\]\[\left| 1/f(x_o) \right| \left| \int\limits\limits\limits\limits_{a}^{b}f''(x)dx \right| =\left| 1/f(x_o) \right| \left| f'(b)-f'(a) \right|\]\[=1/(x_o(1-x_o))\]\[\ge4 \ \ \ since\ (AM \ge GM)\]

Answer 21

Seems all right except ... I don't understand this step.
\[ \int\limits\limits\limits_{a}^{b}\left| f''(x)/f(x) \right|dx \ge \left| 1/f(x_o) \right| \left| \int\limits\limits\limits_{a}^{b}f''(x)dx \right| \]

Answer 22

I guess ... it is right too. since f(x) >= to f(x_0) for all x in (a,b)

Answer 23

\[|1/f(x_o)|\left| \int\limits_{a}^{b} f''(x)dx\right|\]\[=\left| \int\limits\limits\limits_{a}^{b} f''(x)/f(x_o)dx\right|\]\[\le \int\limits\limits\limits\limits_{a}^{b} \left| f''(x)/f(x_o) \right| dx\]\[\le \int\limits\limits\limits\limits_{a}^{b} \left| f''(x)/f(x) \right| dx\ \ \ ,\ \ since\ \ a,b \in(0,1)\]

Answer 24

and since \[f(x)\le f(x_o) \ \ \ ,\forall x \in (0,1)\]