Proof f(x)>g(x) in an interval

[Dank2]

Homework Statement

let fx, gx be continuous in [a,b] and differentiable in (a,b). at the end of the interval f(a) >= g(a).

and f'(x) >g'(x) for a<x<b.

proof f(x) > g(x) for a<x<=b

Attempt:
There is a statement says that if the f'x = g'x for x in [a,b] , then there exists k such that f'x - g'x = k for any x in [a,b]

but f'(x) = g'(x) + t(x), where t(x) isn't have to be a line. and i cannot use the statement

 

[BvU]

and f'(x) > g'(x) for a<x<b.

Shouldn't that be a < sign ?

 

[Dank2]

Shouldn't that be a < sign ?

nope.

 

[BvU]

Might be hard to prove, then:
a=0, b =2, f(x) = x, g(x) = 1 satisifies all criteria, yet f < g on half the interval.

[edit] Ah, the end of the interval is at the beginning. My bad

 

[Dank2]

Might be hard to prove, then:
a=0, b =2, f(x) = x, g(x) = 1 satisifies all criteria, yet f < g on half the interval.

[edit] Ah, the end of the interval is at the beginning. My bad

but f(a) is not equal or or bigger thatn g(a).

 

[BvU]

Yeeah, I get it. Was wrongfooted by the 'end' term.

With f' - g' > 0 and the definition of derivative you should be able to go through the interval without f-g diminishing from its starting value f(a)-g(a).

 

[Dank2]

Yeeah, I get it. Was wrongfooted by the 'end' term.

With f' - g' > 0 and the definition of derivative you should be able to go through the interval without f-g diminishing from its starting value f(a)-g(a).

f'(x) - g'(x) = lim(f(x+h) -f(x) -g(x+h) -g(x))/h > 0

like that? how can i take out f(x) - g(x)

 

[BvU]

Ergo there is a h> 0 such that f(a+h)-g(a+h) > f(a)-g(h)

 

[Dank2]

Ergo there is a h> 0 such that f(a+h)-g(a+h) > f(a)-g(h)

wait but we need x no?, a is the start of the interval

 

[BvU]

So a+h is the new 'beginning' for the next step. Crux is if we can prove this can be continued all through the a,b interval in a finite number of steps.

 

[Dank2]

So a+h is the new 'beginning' for the next step. Crux is if we can prove this can be continued all through the a,b interval in a finite number of steps.

but why doesn't the limit and the h at the denominator making problems here?
what I've got is
lim (f(x+h) - g(x+h))/h > lim (f(x) + g(x))/h

 

[Dank2]

but why doesn't the limit and the h at the denominator making problems here?
what I've got is
lim (f(x+h) - g(x+h))/h > lim (f(x) + g(x))/h

ok i think i got it, the limit doesn't change it much

 

[PeroK]

but why doesn't the limit and the h at the denominator making problems here?
what I've got is
lim (f(x+h) - g(x+h))/h > lim (f(x) + g(x))/h

I'm not sure looking at the definition of the derivative is the best way forward. In any case, it's best to introduce a new function, let's call it $k$ such that:

$k(x) = f(x) - g(x)$

What can you say about $k$? And what theorems do you know?

 

[Dank2]

I'm not sure looking at the definition of the derivative is the best way forward. In any case, it's best to introduce a new function, let's call it $k$ such that:

$k(x) = f(x) - g(x)$

What can you say about $k$? And what theorems do you know?

it is continuous and it is differentiable.
However there is a theorem i know regarding k(x) being constant, but i cannot assume it here.

 

[PeroK]

it is continuous and it is differentiable.
However there is a theorem i know regarding k(x) being constant, but i cannot assume it here.

Do you know the Mean Value Theorem?

 

[Dank2]

yes

 

[PeroK]

yes

It might be useful here!

 

[Dank2]

So;
(f(b) - f(a))/b-a > (g(b)-g(a))/b-a

and i can pick any b-a such that the diffrence is very small, so i can use x instead of it ?? so f(b) - f(a) = f(b)

 

[PeroK]

So;
(f(b) - f(a))/b-a > (g(b)-g(a))/b-a

and i can pick any b-a such that the diffrence is very small, so i can use x instead of it ?? so f(b) - f(a) = f(b)

You need to switch your attention to $k(x)$. Also, I suggest a proof by contradiction using the MVT.

 

[Dank2]

You need to switch your attention to $k(x)$. Also, I suggest a proof by contradiction using the MVT.

ok K'(x) >0 for any x in the interval.

 

[PeroK]

ok K'(x) >0 for any x in the interval.

That's true. What are you trying to prove about $k(x)$?

 

[Mark44]

Thread moved from Precalc section. @Dank2, please post questions involving differentiation (and differentiable functions) in the Calculus & Beyond section.

 

[Dank2]

That's true. What are you trying to prove about $k(x)$?

that is always positive

 

[PeroK]

that is always positive

You need to put a bit more effort in here! You're doing real analysis, which requires precision in terms of the statement of a problem. I'll help you on this point, but you need to start thinking more (pure) mathematically:

We have that $k$ is continuous on $[a, b]$, differentiable on $(a, b)$, $k(a) \ge 0$ and $\forall x \in (a, b), \ k'(x) > 0$.

We must show that $\forall x \in (a, b], \ k(x) > 0$

 

[Dank2]

You need to put a bit more effort in here! You're doing real analysis, which requires precision in terms of the statement of a problem. I'll help you on this point, but you need to start thinking more (pure) mathematically:

We have that $k$ is continuous on $[a, b]$, differentiable on $(a, b)$, $k(a) \ge 0$ and $\forall x \in (a, b), \ k'(x) > 0$.

We must show that $\forall x \in (a, b], \ k(x) > 0$

same problem as before

 

[Dank2]

You need to put a bit more effort in here! You're doing real analysis, which requires precision in terms of the statement of a problem. I'll help you on this point, but you need to start thinking more (pure) mathematically:

We have that $k$ is continuous on $[a, b]$, differentiable on $(a, b)$, $k(a) \ge 0$ and $\forall x \in (a, b), \ k'(x) > 0$.

We must show that $\forall x \in (a, b], \ k(x) > 0$

assume k(x) <0 for some x latter than a, then we will get at one point k'(x) = 0, since at the start of the iterval k(x) > 0 , and it is a continuous function. and that's a contradiction since k'(x) >0 for any x in the interval, and ther fore there is no point where k(x) < 0

 

[PeroK]

assume k(x) <0 for some x latter than a, then we will get at one point k'(x) = 0, since at the start of the iterval k(x) > 0 , and it is a continuous function. and that's a contradiction since k'(x) >0 for any x in the interval, and ther fore there is no point where k(x) < 0

This answer shows that you are lacking mathematical technique. In this case, you need to reserve $x$ as your variable and use, for example, $x_0, x_1, x_2$ for specific values of $x$.

Also, I think you are confusing the Mean Value Theorem with the Intermediate value theorem. Note that we cannot assume that $k'(x)$ is a continuous function, so applying the IVT to $k'(x)$ is not possible. So, you will need to apply the MVT to $k(x)$.

 

[Dank2]

This answer shows that you are lacking mathematical technique. In this case, you need to reserve $x$ as your variable and use, for example, $x_0, x_1, x_2$ for specific values of $x$.

Also, I think you are confusing the Mean Value Theorem with the Intermediate value theorem. Note that we cannot assume that $k'(x)$ is a continuous function, so applying the IVT to $k'(x)$ is not possible. So, you will need to apply the MVT to $k(x)$.

i just get
k'(c) = k(b) - k(a) = f(b)-g(b) - (f(a) - g(a)) (with b-a at the denominator)
or do i not use it right?

 

[PeroK]

i just get
k'(c) = k(b) - k(a) = f(b)-g(b) - (f(a) - g(a)) (with b-a at the denominator)
or do i not use it right?

The MVT applies to any interval. In this case it won't be $[a, b]$. If you find an $x_0$ such that $k(x_0) \le 0$, then you can apply the MVT to $[a, x_0]$.

 

[Dank2]

The MVT applies to any interval. In this case it won't be $[a, b]$. If you find an $x_0$ such that $k(x_0) \le 0$, then you can apply the MVT to $[a, x_0]$.

can't see how to get the contradiction, ill work on it next time, thanks friend.