[prove] monotonicity of function

[Дьявол]

Homework Statement

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

Homework 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)

The Attempt at a Solution

So we need to start with x₁<x₂. Now:
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?

Thanks in advance.

 

[HallsofIvy]

Homework Statement

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

Homework 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)

The Attempt at a Solution

So we need to start with x₁<x₂. Now:
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?

Thanks in advance.

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.

 

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

 

[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

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

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]

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

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

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

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?

Thanks in advance.