Proving Monotonicity for Logarithmic and Fractional Functions?

[zorro]

Homework Statement

I have 2 questions-
1) If x>0, show that log(1+x) < x
2) Is the function f(x) = {x} where {.} denotes fractional part of x increasing on [5,6] ?


A1) My book says that we have to construct a function f(x) = x-log(1+x) ans show that it is increasing and f(0)=0 or positive.
I understand the first point but don't get the second. Why should we prove that f(0) >= 0 ?
0 doesnot even lie in the interval (given in question x>0)

A2) Here f(x) is strictly increasing from [5,6), at x=6, function is not continuous and hence derivative doesnot exist.
So the function is not increasing on [5,6] but on [5,6). Is it correct?

 

[zorro]

Some one help!

 

[╔(σ_σ)╝]

Let f(x) = x-log(1+x)

The way I see it you have to show that f(0)=0 or positive because you are told to prove that for x>0, show that log(1+x) < x . This should certainly hold true for x very very close to 0.

A2) Here f(x) is strictly increasing from [5,6), at x=6, function is not continuous and hence derivative doesnot exist.
So the function is not increasing on [5,6] but on [5,6). Is it correct?

What do you mean by fractional ? Do you mean decimal portion of the number ?

Either way your resononing is not correct.

The function is actually increase.

Consider the function x-5

f(x)= {x} = x-5 on [5,6)

This function is continuous and has a nice continuous derievative and is strictly increasing on [5, 6).

 

[zorro]

Let f(x) = x-log(1+x)

What do you mean by fractional ? Do you mean decimal portion of the number ?

Either way your resononing is not correct.

The function is actually increase.

Consider the function x-5

f(x)= {x} = x-5 on [5,6)

This function is continuous and has a nice continuous derievative and is strictly increasing on [5, 6).

The answer given is that it increases on the interval (5,6) and not [5,6)

 

[╔(σ_σ)╝]

I am not sure why that is. Did you copy down the problem correctly?

 

[zorro]

I am not sure why that is. Did you copy down the problem correctly?

Yes. I rechecked it.

 

[╔(σ_σ)╝]

Well it makes sense. Since f ' is defined of (5, 6) and since we do not "know" the values of f before 5 we cannot talk about it increasing. Actually we know the values or f before 5 and it turns out that at x=5 the function is actually decreasing since you have points like 4. 9991.., 4. 999999997... etc.

I was not careful in my thinking.

 

[zorro]

f ' is defined on [5,6)
I don't understand what you mean by not knowing the values of f before 5. Why do we need to consider them? We don't find whether a function is decreasing /increasing at a point, how did you infer that the function decreases at x=5?

 

[micromass]

Can you define "fractional part" for me?

 

[zorro]

Fractional part function is defined as f(x) = x - [x] where [.] denotes greatest integer less than or equal to x ( for all x belonging to R )

You can refer this for more detail:
http://mathworld.wolfram.com/FractionalPart.html

 

[micromass]

Then for (2), it is correct that the function increases on [5,6[, but not when you add 6 to that interval...

 

[zorro]

So you mean the answer given ]5,6[ is wrong?

 

[micromass]

I think so, yes...

 

[zorro]

Ok.
Thank you so much!