Can I use the mean value theorem to prove that f>g for all x in (a,b)?

[Dank2]

Assume f and g are two continuous functions in (a, b).
If at the start of the segment I've shown f>g by taking the lim where x ---> a+ and the f ' > g ' for every x in (a,b )
can i say that f >g for all x in (a,b )? is there a theorem for that? that looks intuitively right.

 

[LCKurtz]

You need to state more clearly what you are assuming.

Assume f and g are two continuous functions in (a, b).
If at the start of the segment I've shown f>g by taking the lim where x ---> a+ and the f ' > g ' for every x in (a,b )
can i say that f >g for all x in (a,b )? is there a theorem for that? that looks intuitively right.

You say $f' > g'$ but you only assumed $f$ and $g$ are continuous on $(a,b)$. Do you mean to assume they are differentiable? And are you saying that $\lim_{x \to a^+} f(x) > \lim_{x\to a^+}g(x)$? If the answers are yes to these, I suggest you try proving your result using the mean value theorem.

 

[Dank2]

yes, they are differentiable.
https://www.physicsforums.com/threads/proof-inx-x-1.900384/#post-5667094
i proved it in other way in the link above post 9.
howcan i use mean value theorem to show it ?

 

[LCKurtz]

Define $f(a)$ and $g(a)$ equal to their right hand limits at $a$ so you can use $[a,b)$ and $f(a)>g(a)$. Let $h(x) = f(x) - g(x)$, so you have $h'(x)>0$. What does the mean value theorem tell you about $h$ if you have a point $c$ in $(a,b)$ where $f(c)<g(c)$?

 

[Dank2]

Define $f(a)$ and $g(a)$ equal to their right hand limits at $a$ so you can use $[a,b)$ and $f(a)>g(a)$. Let $h(x) = f(x) - g(x)$, so you have $h'(x)>0$. What does the mean value theorem tell you about $h$ if you have a point $c$ in $(a,b)$ where $f(c)<g(c)$?

if f(c)<g(c) then h(c) < 0, doesn't help me much

 

[LCKurtz]

I asked you what the mean value theorem tells you. Write it down.

 

[Dank2]

That there is point c where the derivative of h'(c) is parallel to the straight line connecting the two end points of the segment

 

[LCKurtz]

You need to write down the equation the mean value theorem tells you applied to $h$.

 

[Dank2]

H'(c)= (h(b)-h(a))/b-a > 0

 

[LCKurtz]

H'(c)= (h(b)-h(a))/b-a > 0

Forget the > 0 for a minute. Examine the sign on the left side versus the sign on the right side.
[Edit, added] I overlooked you haven't used the point c properly on the right side. You don't know anything about h(b). And you don't want c on the left side.

 

[LCKurtz]

Now that the football game I was watching while I was working with you is over, I will be a bit more explicit about what you are doing wrong. Here is what I suggested in post #5:

Define $f(a)$ and $g(a)$ equal to their right hand limits at $a$ so you can use $[a,b)$ and $f(a)>g(a)$. Let $h(x) = f(x) - g(x)$, so you have $h'(x)>0$. What does the mean value theorem tell you about $h$ if you have a point $c$ in $(a,b)$ where $f(c)<g(c)$?

Then I asked you to write what the mean value theorem tells you about $h$. You wrote this:

H'(c)= (h(b)-h(a))/(b-a) > 0

There are several things wrong with this besides missing parentheses. First, there is no > 0 in the mean value theorem. Second, you can't use $c$ in $H'(c)$ because I have already used it above where you assume $c$ is a point where $f(c)<g(c)$. Third, $f$ and $g$ aren't defined at $b$ so neither is $h(b)$.
What you need to do is apply the mean value theorem to $h$ on the interval $[a,c]$. Write it out carefully and completely. You will need to use the words "there exists" somewhere in your application of the mean value theorem.