[prove] monotonicity of function

  1. 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

    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.
     
  2. jcsd
  3. HallsofIvy

    HallsofIvy 40,689
    Staff Emeritus
    Science Advisor

    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.
     
  4. 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]
     
  5. 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 :smile:

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

    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.
     
  6. 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.
     
  7. 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.
     
    Last edited: Dec 7, 2008
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