[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&lt;x_2 ; x_1-1&lt;x_2-1 ; \frac{x_1-1}{x_2-1}&lt;\frac{x_2-1}{x_2-1} ; \frac{x_1-1}{x_2-1}&lt;1 ; ln\frac{x_1-1}{x_2-1}&lt;ln(1) ; ln\frac{x_1-1}{x_2-1}&lt;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&lt;x_2 ; x_1-1&lt;x_2-1 ; \frac{x_1-1}{x_2-1}&lt;\frac{x_2-1}{x_2-1} ; \frac{x_1-1}{x_2-1}&lt;1 ; ln\frac{x_1-1}{x_2-1}&lt;ln(1) ; ln\frac{x_1-1}{x_2-1}&lt;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&lt; x_2, progress to ln\frac{x_1-1}{x_2-1}&lt; 0 and then assert that ln(x_1- 1)&lt; ln(x_2-1).
Notice, by the way, that you need x_1&gt;1 and x_2&gt; 1 in order to assert that x_1-1&gt; 0, x_2- 1&gt; 0 so \frac{x_2-1}{x_1-1}&gt; 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&lt;\frac{x_1-x_2}{x_2-1}&lt;0

0&lt;1+\frac{x_1-x_2}{x_2-1}&lt;1

and out of here ln(1+\frac{x_1-x_2}{x_2-1})&lt;ln(1)

ln(1+\frac{x_1-x_2}{x_2-1})&lt;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&lt;\frac{x_1-1}{x_2-1}&lt;1

and out of here ln\frac{x_1-1}{x_2-1}&lt;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&gt;x_1 ; sin(x_2)&gt;sin(x_1) ; sin(x_2)-sin(x_1)&gt;0

or sin(x_1)&gt;0 ; -sin(x_1)&lt;0 ; 1-sin(x_1)&lt;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.