Application of the mean value theorem.

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Homework Help Overview

The discussion revolves around the application of the Mean Value Theorem to two continuous functions, f and g, defined on the interval [a,b]. The problem requires showing that the difference in the function values, f(b) - f(a), is less than the difference in the values of g, g(b) - g(a), given that the derivative of f is always less than that of g on the open interval (a,b).

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the Mean Value Theorem, discussing the relationship between the derivatives of the functions and the differences in their values. There is a focus on the validity of dividing the derivatives and the assumptions regarding g'(c).

Discussion Status

The discussion is ongoing, with participants questioning the assumptions made about g'(c) and the method of division used in the original attempt. Some participants express uncertainty about the necessity of proving certain assumptions, while others provide examples to illustrate their points.

Contextual Notes

Participants note the potential danger of division by zero if g'(c) equals zero, raising concerns about the assumptions made in the problem. There is an example provided to clarify the behavior of the functions and their derivatives within the specified interval.

mtayab1994
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Homework Statement



Let f and g be two continuous functions on [a,b] and differentiable on ]a,b[ such that for every x in ]a,b[ : f'(x)<g'(x)

Homework Equations



Show that f(b)-f(a)<g(b)-g(a)

The Attempt at a Solution



So I said that there exists a c in ]a,b[ such that f'(c)=(f(b)-f(a))/(b-a) and g'(c)=(g(b)-g(a))/(b-a) and since f'(x)<g'(x) then f'(c)<g'(c) and that implies that f'(c)/g'(c)<1 and when substituting in what we found we get that (f(b)-f(a))/(g(b)-g(a))<1 and that implies that f(b)-f(a)<g(b)-g(a). Is this correct??
 
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Yes, that is correct, but I don't understand why you did the division (also this is dangerous- what if [itex]g'(c) =0[/itex]). Do you see how you can do it without the division?
 
Robert1986 said:
Yes, that is correct, but I don't understand why you did the division (also this is dangerous- what if [itex]g'(c) =0[/itex]). Do you see how you can do it without the division?

No I can't see how, but i can suppose that g'(c)≠0 right?
 
mtayab1994 said:
No I can't see how, but i can suppose that g'(c)≠0 right?
I don't see how you can suppose that (I might be missing something, but if you can, you will have to prove why it is true, which isn't necessary.) For example, if [itex]f(x)=-5x[/itex] and [itex]g(x)=-x^2+1[/itex] and [itex]a=-1[/itex] and [itex]b=1[/itex] the smallest that [itex]g'[/itex] gets is -2. But, [itex]f'(x)=-5 < -2 < g'(x)[/itex]. Right?

So, you know that:
[itex]f'(c) < g'(c)[/itex] and
[itex]f'(c)=\frac{f(b)-f(a)}{b-a}[/itex] and [itex]g'(c)=\frac{g(b)-g(a)}{b-a}[/itex].

Now do you see?
 
Robert1986 said:
I don't see how you can suppose that (I might be missing something, but if you can, you will have to prove why it is true, which isn't necessary.) For example, if [itex]f(x)=-5x[/itex] and [itex]g(x)=-x^2+1[/itex] and [itex]a=-1[/itex] and [itex]b=1[/itex] the smallest that [itex]g'[/itex] gets is -2. But, [itex]f'(x)=-5 < -2 < g'(x)[/itex]. Right?

So, you know that:
[itex]f'(c) < g'(c)[/itex] and
[itex]f'(c)=\frac{f(b)-f(a)}{b-a}[/itex] and [itex]g'(c)=\frac{g(b)-g(a)}{b-a}[/itex].

Now do you see?

So all you are going to do now is say that since f'(c)<g'(c) and when we cancel the b-a we will get f(b)-f(a)<g(b)-g(a) right??
 
Yep!
 
Robert1986 said:
Yep!

Alright thanx!
 

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