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[prove] monotonicity of function

  1. Dec 6, 2008 #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. Dec 6, 2008 #2

    HallsofIvy

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    Staff Emeritus
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    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. Dec 6, 2008 #3
    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. Dec 6, 2008 #4
    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. Dec 6, 2008 #5
    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. Dec 7, 2008 #6
    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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