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 ⊆ D_{f} 2. Relevant equations a) monotonically increasing if the set E ⊆ D_{f} for arbitrary numbers x_{1}, x_{2} ∊ E and x_{1}<x_{2} ⇒ f(x_{1})<f(x_{2}) b)monotonically decreasing if the set E ⊆ D_{f} for arbitrary numbers x_{1}, x_{2} ∊ E and x_{1}<x_{2} ⇒ f(x_{1})>f(x_{2}) 3. The attempt at a solution So we need to start with x_{1}<x_{2}. Now: f(x_{1})-f(x_{2})=ln(x_{1}-1)-ln(x_{2}-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(x_{1})-f(x_{2})<0 and f(x_{1})<f(x_{2}) 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 [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.
Thanks for the post. Yes, I see now that I missed the fact that x_{1}>1 and x_{2}>1 so I could used that fact that x_{2}-1>0 and x_{1}-x_{2}<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: x_{1}-1>0 and x_{2}-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]
And what if I have f(x)=3^{-x} ? x_{1}<x_{2} f(x_{1})-f(x_{2})=3^{-x1} - 3^{-x2}=1/3^{x1} - 1/3^{x2}= Now let's try with LaTeX =[tex]\frac{3^{x_1}-3^{x_2}}{3^{x_1+x_2}}[/tex] Out of x_{1}<x_{2} log_{3}3^{x1}<log_{3}3^{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(x_{1})-f(x_{2})<0 and f(x_{1})<f(x_{2}) so that the function is monotonically increasing.
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.
Can somebody please help with f(x)=x-sin(x), E=(0,п) x_{1}<x_{2} [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. x_{1}-x_{2}<0 and if п>x>0 then 1>sinx>0. If 1>sin(x_{2})>0 ; 1-sin(x_{1})>sin(x_{2})-sin(x_{1})>-sin(x_{1)} 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(x_{2})-sin(x_{1})<0 and x_{1}-x_{2}<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.