[prove] monotonicity of function

Дьявол

1. The problem statement, all variables and given/known data
Show that the function is monotonic, and if so find if it increases or decreases monotonically.

f(x)=ln(x-1), E=(1,∞) where E ⊆ D_f


2. Relevant equations

a) monotonically increasing if the set E ⊆ D_f for arbitrary numbers x₁, x₂ ∊ E and x₁<x₂ ⇒ f(x₁)<f(x₂)

b)monotonically decreasing if the set E ⊆ D_f for arbitrary numbers x₁, x₂ ∊ E and x₁<x₂ ⇒ f(x₁)>f(x₂)

3. The attempt at a solution

So we need to start with x₁<x₂. Now:
f(x₁)-f(x₂)=ln(x₁-1)-ln(x₂-1)=
=[tex]ln\frac{x_1-1}{x_2-1}=ln\frac{x_2-1+x_1-x_2}{x_2-1}=ln(1+\frac{x_1-x_2}{x_2-1})[/tex]
But I am stuck in here proving, so I tried:
[tex]x_1<x_2[/tex] ; [tex]x_1-1<x_2-1[/tex] ; [tex]\frac{x_1-1}{x_2-1}<\frac{x_2-1}{x_2-1}[/tex] ; [tex]\frac{x_1-1}{x_2-1}<1[/tex] ; [tex]ln\frac{x_1-1}{x_2-1}<ln(1)[/tex] ; [tex]ln\frac{x_1-1}{x_2-1}<0[/tex]
so f(x₁)-f(x₂)<0 and f(x₁)<f(x₂) and the function is monotonically increasing. Is this correct? Can I always use this method?

Thanks in advance.

HallsofIvy

It would be simplest to show that the derivative is always positive or always negative for [itex]x\ge 1[/itex]. Assuming that you are not allowed to or cannot take the derivative what you have done is perfectly valid- and very good! A correct proof, of course, would start with [itex]x_1< x_2[/itex], progress to [itex]ln\frac{x_1-1}{x_2-1}< 0[/itex] and then assert that [itex]ln(x_1- 1)< ln(x_2-1)[/itex].
Notice, by the way, that you need [itex]x_1>1[/itex] and [itex]x_2> 1[/itex] in order to assert that [itex]x_1-1> 0[/itex], [itex]x_2- 1> 0[/itex] so [itex]\frac{x_2-1}{x_1-1}> 0[/itex] and [itex]ln\frac{x_2-1}{x_1-1}[/itex] exists.

Дьявол

Thanks for the post. Yes, I see now that I missed the fact that x₁>1 and x₂>1 so I could used that fact that x₂-1>0 and x₁-x₂<0 so that out of here:

[tex]ln(1+\frac{x_1-x_2}{x_2-1})[/tex]

[tex]-1<\frac{x_1-x_2}{x_2-1}<0[/tex]

[tex]0<1+\frac{x_1-x_2}{x_2-1}<1[/tex]

and out of here [tex]ln(1+\frac{x_1-x_2}{x_2-1})<ln(1)[/tex]

[tex]ln(1+\frac{x_1-x_2}{x_2-1})<0[/tex]

or it would be much simple if I did:

x₁-1>0 and x₂-1>0

and

[tex]ln\frac{x_1-1}{x_2-1}[/tex]

so that

[tex]0<\frac{x_1-1}{x_2-1}<1[/tex]

and out of here [tex]ln\frac{x_1-1}{x_2-1}<0[/tex]

Дьявол

And what if I have f(x)=3^-x ?

x₁<x₂

f(x₁)-f(x₂)=3^-x₁ - 3^-x₂=1/3^x₁ - 1/3^x₂=

Now let's try with LaTeX

=[tex]\frac{3^{x_1}-3^{x_2}}{3^{x_1+x_2}}[/tex]

Out of x₁<x₂
log₃3^x₁<log₃3^x₂

Can I use this method to prove?

Now 3^x₁<3^x₂

But how to prove 3^{x₁ + x₂}>0 ? Or it doesn't need proving?

Now it turns out that f(x₁)-f(x₂)<0 and f(x₁)<f(x₂) so that the function is monotonically increasing.

mutton

It seems obvious that any power with a positive base is positive (like the multiplication of positive numbers), but you could always write this down.

Дьявол

Can somebody please help with f(x)=x-sin(x), E=(0,п)
x₁<x₂

[tex]f(x_1)-f(x_2)=x_1-sin(x_1)-x_2+sin(x_2)=x_1-x_2+sin(x_2)-sin(x_1)[/tex]

I am stuck in here. x₁-x₂<0 and if п>x>0 then 1>sinx>0.

If 1>sin(x₂)>0 ; 1-sin(x₁)>sin(x₂)-sin(x₁)>-sin(x₁₎

If [tex]x_2>x_1 ; sin(x_2)>sin(x_1) ; sin(x_2)-sin(x_1)>0[/tex]

or [tex]sin(x_1)>0 ; -sin(x_1)<0 ; 1-sin(x_1)<1[/tex] so,

1<sin(x₂)-sin(x₁)<0 and x₁-x₂<0

I got plenty of information but I don't know if I am using it correctly.

What should I do now?

Thanks in advance.