f(x) continues in borders [a,b], derivitable in borders (a,b

transgalactic said:
the question: there is a function f(x) which is continues in the borders [a,b] and derivitable in the borders (a,b), b>a>0 alpha differs 0
Click to expand...
Do you mean that the function is "continous" on the closed "interval", and "differentiable" on the open "interval"...? What is "alpha", and that does it mean for "alpha differs zero"?

transgalactic said:
prove the there is b>c>a
Click to expand...
"Prove the ["that"?] there is b > c > a" such that... what? How does "c" relate?

transgalactic said:
in that formula:....
Click to expand...
Image not viewable. Sorry.

Eliz.
 
derivitable?. I like that one. That is the malaprop of the month. :D
 
f(x) continues in [a,b] interval,and differentiable (a,b) b>a>0
alpha differs 0
proove that there is b>c>a

in that formula:
http://img392.imageshack.us/my.php?image=81208753je3.gif

my trial:
i mark alpha as "&"
mean theorim says f'(c)=[(f(b)-f(a)]/(b-a)

[f(b)*(a^&) - f(a)*b^&] / [a^& - b^&]= f(c) -c * [(f(b)-f(a)]/[(b-a) * &]

using mean theorem i replaced f'(c)
what should i do next??
 
How 'bout you actually read Halls' post at Physics forums instead of reverting to repeating the same post arbitrarily all of the internet. I presume you won't have a laptop and internet connection in your exams.
 
f(x) continues in [a,b] interval,and differentiable (a,b) b>a>0
alpha differs 0
proove that there is b>c>a

in that formula:
http://img392.imageshack.us/my.php?image=81208753je3.gif

solution:
alpha=&
i was told that in this question we use
the "couchy mean theorim"
[f(b)-f(a)]/[g(b)-g(a)]=f'(c)/g'(c)

F(x)=f(x)/[x^&] G(x)=1/(x^&)

[F(b)-F(a)]/[G(b)-G(a)] = [f(b)/(b^&) - f(a)/(a^&)]/[1/(b^&) - 1/(a^&)] =

= [f(b)*(a^&) - f(a)*( b^&)]/[a^& - b^&]

F'(c) / G'(c)=[f'(c)* (1/c^&) -&*c^(-&-1)*f(c)]/[-&*c^(-&-1)]=

=[&*f(c)-c*f'(c)]/& =f(c) - c*f'(c)/&

we need to use couchy mean theorim
i dont know how they desided to pick this values for F(x) and G(x)??

what proccess do i need to do in order to deside
that the way of solving such a problem is picking

F(x)=f(x)/[x^&] G(x)=1/(x^&)


??
 
Third post of question merged with previous two threads.
 
Unco said:
How 'bout you actually read Halls' post at Physics forums instead of reverting to repeating the same post arbitrarily all of the internet. I presume you won't have a laptop and internet connection in your exams.
Click to expand...

i read this solution
i cant
the main problem in this sort of questions
is choosing the F(X) and G(x)

i got a very simple solution to this question using cauchy mean theorem

but i dont know how to choose F(x) and G(x)

??
 
You must log in or register to reply here.
Top