Regarding the previusmean theorim question

[transgalactic]

i got a different solution to my problem :

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 i don't use "mean theorim"
but 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)/&

my proffesor solved it like this after about 50 minutes when
he tried to use the mean theorim
and then he said that we need to use couchy mean theorim
i don't know how he desided that??

and the second most important problem is picking the variables.
how he picked
F(x)=f(x)/[x^&] G(x)=1/(x^&)
??
what process 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^&)


?

 

[mighty2000]

Thank you for sharing your solution to the problem. It seems like you have used the Cauchy Mean Theorem correctly to prove the existence of a point c between a and b. To address your questions, I will first explain why your professor may have chosen to use the Cauchy Mean Theorem instead of the Mean Value Theorem.

The Mean Value Theorem states that if a function f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point c between a and b such that f'(c) = [f(b)-f(a)]/(b-a). This theorem is commonly used to prove the existence of a point c where the derivative of a function is equal to its average rate of change over the interval [a,b]. However, in the problem you presented, the function g(x) = 1/(x^&) is not differentiable at x = 0. Therefore, the Mean Value Theorem cannot be applied.

In this case, the Cauchy Mean Theorem is a better choice because it does not require the function to be differentiable at the endpoints of the interval. It states that if two functions f and g are continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point c between a and b such that [f(b)-f(a)]/[g(b)-g(a)] = f'(c)/g'(c). This theorem allows us to prove the existence of a point c where the ratio of the derivatives of f and g is equal to the ratio of their average rates of change over the interval [a,b].

As for your question about how your professor chose the variables F(x) = f(x)/[x^&] and G(x) = 1/(x^&), it is likely that he recognized that these functions satisfy the conditions of the Cauchy Mean Theorem. In order to use the theorem, we must have two continuous functions on the closed interval [a,b] and differentiable functions on the open interval (a,b). By choosing these functions, your professor was able to prove the desired result using the Cauchy Mean Theorem.

I hope this helps to clarify the reasoning behind your professor's solution. Keep up the good work in your studies of mathematics and science!