# [prove] monotonicity of function

1. Dec 6, 2008

### 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 ⊆ Df

2. Relevant equations

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

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

3. The attempt at a solution

f(x1)-f(x2)=ln(x1-1)-ln(x2-1)=
=$$ln\frac{x_1-1}{x_2-1}=ln\frac{x_2-1+x_1-x_2}{x_2-1}=ln(1+\frac{x_1-x_2}{x_2-1})$$
But I am stuck in here proving, so I tried:
$$x_1<x_2$$ ; $$x_1-1<x_2-1$$ ; $$\frac{x_1-1}{x_2-1}<\frac{x_2-1}{x_2-1}$$ ; $$\frac{x_1-1}{x_2-1}<1$$ ; $$ln\frac{x_1-1}{x_2-1}<ln(1)$$ ; $$ln\frac{x_1-1}{x_2-1}<0$$
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?

2. Dec 6, 2008

### HallsofIvy

Staff Emeritus
It would be simplest to show that the derivative is always positive or always negative for $x\ge 1$. 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 $x_1< x_2$, progress to $ln\frac{x_1-1}{x_2-1}< 0$ and then assert that $ln(x_1- 1)< ln(x_2-1)$.
Notice, by the way, that you need $x_1>1$ and $x_2> 1$ in order to assert that $x_1-1> 0$, $x_2- 1> 0$ so $\frac{x_2-1}{x_1-1}> 0$ and $ln\frac{x_2-1}{x_1-1}$ exists.

3. Dec 6, 2008

### 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:

$$ln(1+\frac{x_1-x_2}{x_2-1})$$

$$-1<\frac{x_1-x_2}{x_2-1}<0$$

$$0<1+\frac{x_1-x_2}{x_2-1}<1$$

and out of here $$ln(1+\frac{x_1-x_2}{x_2-1})<ln(1)$$

$$ln(1+\frac{x_1-x_2}{x_2-1})<0$$

or it would be much simple if I did:

x1-1>0 and x2-1>0

and

$$ln\frac{x_1-1}{x_2-1}$$

so that

$$0<\frac{x_1-1}{x_2-1}<1$$

and out of here $$ln\frac{x_1-1}{x_2-1}<0$$

4. Dec 6, 2008

### Дьявол

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

x1<x2

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

Now let's try with LaTeX

=$$\frac{3^{x_1}-3^{x_2}}{3^{x_1+x_2}}$$

Out of x1<x2
log33x1<log33x2

Can I use this method to prove?

Now 3x1<3x2

But how to prove 3x1 + 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.

5. Dec 6, 2008

### 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.

6. Dec 7, 2008

### Дьявол

x1<x2

$$f(x_1)-f(x_2)=x_1-sin(x_1)-x_2+sin(x_2)=x_1-x_2+sin(x_2)-sin(x_1)$$

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 $$x_2>x_1 ; sin(x_2)>sin(x_1) ; sin(x_2)-sin(x_1)>0$$

or $$sin(x_1)>0 ; -sin(x_1)<0 ; 1-sin(x_1)<1$$ 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?