How to Prove that g(x) is Increasing on [0, Infinity) and f(x) = 0 on [0, c]?

[inferi]

hi,

this question is so hard that i do not know it's head from it's tail.
here this the question:
let f be a funcation defined on [o,infinity),with the properties: f is continuous on [o,infinity),f(0)=0,and the first derivative of f exists on [o,infinity),and the first derivative of f increasing on [o,infinity),

1-show that g(x)=(f(x)/x) is increasing on [o,infinity).


2-show that if the first derivative of f(c)=0 for some c>0, if f(x)>=0,then f(x)=0 on the inteval [0,c].

please i need a big fat hint for this question anyone can help? thank you

 

[JasonRox]

Did you find the derivative of g(x) yet?

 

[inferi]

so i took the derivative of g(x) and it is equal to =(x.derivative of f(x)-f(x))/x^2
then you equal it to zero (first derivative test) so the final answer is
derivative of f(x)=f(x)/x
and from the question we know that derivative of f(x) is continuous on [o,infinity) so f(x)/x (which is g(x)) is also continuous on [o,infinity) .
is that the right answer for part 1 if it is tell me and give a hint for par 2.
thank you

 

[jbowers9]

Perturbed Theory

Hi all. Attached is a pdf file w/ the appropriate refs.
I've been reading a paper at the following link:

www.cims.nyu.edu/~eve2/reg_pert.pdf

I have several questions:

In the first example they use the method to approximate the roots for

x^2 - 1 = "epsilon" x

I was under the impression - wrongly perhaps - that f(x) had to have continuous derivatives to use the Taylor series. When you go to f'''(a), the coefficient c3 vanishes. And, strictly speaking isn't this a Maclaurin series?

In the problem for x^2 -1 = "epsilon" e^x
The coefficient X2 for x = -1 is 3/8e^2.
I'm sure of this since I went through it several times.

I don't understand what the second figure - Figure 2 - is referring to at all (why two)??
And what "3" solutions? Doesn't a quadratic have 2 solutions? What is the solid line in the figure a plot of anyway?

I apologize that I can't post it verbatim but I'm unable to cut and paste from an Adobe file to this format and I'm really having problems using the LaTex editor. I'd appreciate any commentary. Thank you.
John

 

[inferi]

you John or whatever this topic is for the question in the first post so be good and get out and make your own topic ok

 

[cristo]

I think what inferi means is that if you have a new questio, John, you should start a new thread about it in the appropriate forum.

Both of you should note that homework/textbook questions should be posted in the homework forums and not the main maths forums. Please do so in future.

 

[EnumaElish]

I was under the impression - wrongly perhaps - that f(x) had to have continuous derivatives to use the Taylor series. When you go to f'''(a), the coefficient c3 vanishes.

The function F(x;ε) = f(x) - εe^x in your example has continuous derivatives (w/r/t x).

And, strictly speaking isn't this a Maclaurin series?

My understanding is, it will be a Macl. series if the derivatives are evaluated at x = 0.

In the problem for x^2 -1 = "epsilon" e^x
The coefficient X2 for x = -1 is 3/8e^2.
I'm sure of this since I went through it several times.

Is Fig. 2 meant for F, or one of the roots of F(x) = 0, which would be x^*(ε)?

I don't understand what the second figure - Figure 2 - is referring to at all (why two)??
And what "3" solutions? Doesn't a quadratic have 2 solutions?

f is quadratic, F isn't.

What is the solid line in the figure a plot of anyway?

One of the roots of F = 0, as a function of ε.

 

[al-mahed]

I tought in e:

d/dx (e^u) = e^u *du

what is the n-th derivative of e^kx? ( k = constant and x = variable)

f = e^kx - 1 obs: f(0) = 0
f'= ke^kx
f''=k^2*e^kx
(...)

the n-th derivative = k^n*e^kx

g = f/x = (e^kx - 1)/x

e^kx - 1 is increasing faster than x, because f' > g', so g will grow when x grows

but I couldn't satisfy the second condition

 

[inferi]

so what about the second part of my question? please anyone can help?

 

[Office_Shredder]

al-mahed, you tried solving it for a specific f, not for all possible functions.

If $$f(x)=(x-1)^2-1$$. f(x) is continuous on [0,infinity), the first derivative exists on the interval, the first derivative is increasing, f(0)=0 yet the second property is not satisfied. Similiarly, f(x)=(x+1)^2-1 doesn't satisfy the first property. Is there a condition on what the derivative of f at 0 is or something like that?

 

[SiddharthM]

This is another problem from rudin and is the first nontrivial question on the Differentiation problem set. Play with the facts given - after a couple of hours of struggling you'll know it's head from it's tail. Then maybe after a couple of days you'll stumble upon the correct solution - posting here for answers is NOT the point of taking an analysis course.

You will need to reread and absorb the theorems on differentiation to solve this. Judging from your last post you don't understand what the mean value theorem is saying and without that understanding the solution will remain elusive.

 

[inferi]

This is another problem from rudin and is the first nontrivial question on the Differentiation problem set. Play with the facts given - after a couple of hours of struggling you'll know it's head from it's tail. Then maybe after a couple of days you'll stumble upon the correct solution - posting here for answers is NOT the point of taking an analysis course.

You will need to reread and absorb the theorems on differentiation to solve this. Judging from your last post you don't understand what the mean value theorem is saying and without that understanding the solution will remain elusive.

if tou do not know to answer just say so, but do not be like someone that thinks he is better than anyone ok.

 

[SiddharthM]

if tou do not know to answer just say so, but do not be like someone that thinks he is better than anyone ok.

First of all, I DO know a solution. When I took my intro analysis course over a year ago this was a problem I worked on, I struggled with it for days before I got the solution, but i DID get it, and on my own at that.

You have posted the question on a forum without any proof that you have even ATTEMPTED a proof. On your last thread you did the same thing, moreover you don't seem to understand what the theorems are saying. You are learning the subject for the first time, it's a bad idea to ask for "big fat hints" whenever a solution doesn't come to you right away.

I'm simply telling you that you need to WORK if you want to learn analysis, the purpose of this board is to HELP but that doesn't mean we'll do the work for you.

 

[SiddharthM]

so i took the derivative of g(x) and it is equal to =(x.derivative of f(x)-f(x))/x^2
then you equal it to zero (first derivative test) so the final answer is
derivative of f(x)=f(x)/x
and from the question we know that derivative of f(x) is continuous on [o,infinity) so f(x)/x (which is g(x)) is also continuous on [o,infinity) .
is that the right answer for part 1 if it is tell me and give a hint for par 2.
thank you

useful notation: the derivative of f(x)=f '(x)

so you are correct when you say g '(x)=(xf '(x)-f(x))/x^2. Why are you setting it to zero?

The hypothesis does NOT say that f '(x) is continuous on [0, infinity), but it does say it EXISTS and is increasing on that interval. Even if f(x)/x is continuous what is your argument that shows that this fact implies g(x) is increasing?

Query: What fact about g'(x) implies that g(x) is increasing?
Query: What is the exact statement of the mean value theorem.
Query: Under the given conditions of the problem, where can you apply the mean value theorem?

 

[JonF]

For #1 I would just take the derivative in a straight forward manner using the quotient rule. For #2 it seems to be pretty straight forward if you try a “proof by contradiction”.

 

[al-mahed]

Hi SiddharthM,

Thx for your help here!

I never took analysis classes, but i´ll try to do something...

[1] mean value theorem: if f(x) is continuous on [a,b] and L is a real number such that f(a)<L<f(b) or f(b)<L<f(a), then there is at least a "c" on [a,b] such that f(c)=L


Query: What fact about g'(x) implies that g(x) is increasing?

A: g'(x)>0

Query: What is the exact statement of the mean value theorem.

A: [1]

Query: Under the given conditions of the problem, where can you apply the mean value theorem

A: to prove that f(x)=0 on the inteval [0,c]

We need to prove f(x) is continuous on [0,c] in order to apply the mean value theorem

 

[SiddharthM]

you haven't taken an analysis course? advanced calculus? Frankly you are in over your head if you have never taken or not even enrolled in one of these courses.

[1] is the intermediate value theorem not the mean value theorem.

yes, if you show g'(x) is greater than (or equal) to zero on a interval then it follows that g(x) is increasing on that interval.

 

[al-mahed]

Analysis = calculus? Or saying analysis you mean Real Analysis?

Well, you said that if g'(x) = 0 g(x) is increasing? correct?

[1]g(x)=5 ==> g'(x) = 0 ==> there is no inclination of the tangent at all

[2]g(x) = x^2 - 2x ==> g(x) = 2x - 2 ==> there is no inclination on x = 1

in [1] g(x) never increases

in [2] g(x) do not increase if x = 2

ps: what means "you are in over your heard"? (I'm not so good in english)

 

[al-mahed]

bt the way, could someone state correctly the problem?

 

[SiddharthM]

Analysis is not calculus, it is sometimes called advanced calculus and sometimes real analysis. Although Real analysis can also refer to higher level analysis - measure theory. Analysis is basically a rigorous treatment of Calculus in which one builds the theory of real numbers so that the theorems of calculus can be proven.

The problem is correctly stated in the first post of the thread.

If a function is monotonically increasing it means that if x<y then f(x) (less than or equal to) f(y). So the constant functions ARE monotonically increasing (and monotonically decreasing for that matter). The definition for a Strictly increasing function means the inequality is strict (ie < or > but not equal to).

I say in over your head to say that the problem requires a deeper understanding than that acquired at the level of calculus.

 

[al-mahed]

From: http://en.wikipedia.org/wiki/Mean_value_theorem

In calculus, the mean value theorem states, roughly, that given a section of a smooth curve, there is a point on that section at which the derivative (slope) of the curve is equal (parallel) to the "average" derivative of the section. It is used to prove theorems that make global conclusions about a function on an interval starting from local hypotheses about derivatives at points of the interval.

This theorem can be understood concretely by applying it to motion: if a car travels one hundred miles in one hour, so that its average speed during that time was 100 miles per hour, then at some time its instantaneous speed must have been exactly 100 miles per hour.
the correct mean value theorem:

Let f : [a, b] → R be a continuous function on the closed interval [a, b], and differentiable on the open interval (a, b). Then, there exists some c in (a, b) such that

$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

 

[therector24]

for joen how can u make through contradicion give an example, and for SiddharthM the critical point for g'(x) is f'(x) and through the first derivative test it can be proved i imagined that if you can solve the first part so you will do it in the second part but i don't know how.

 

[therector24]

no one can wants to give a hand on this !(even a word )