[prove] monotonicity of functionby Дьявол Tags: decreasing, monotonic increasing, monotonicity, proving 

#1
Dec608, 05:17 AM

P: 365

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(x1), 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_11}{x_21}=ln\frac{x_21+x_1x_2}{x_21}=ln(1+\frac{x_1x_2}{x_21})[/tex] But I am stuck in here proving, so I tried: [tex]x_1<x_2[/tex] ; [tex]x_11<x_21[/tex] ; [tex]\frac{x_11}{x_21}<\frac{x_21}{x_21}[/tex] ; [tex]\frac{x_11}{x_21}<1[/tex] ; [tex]ln\frac{x_11}{x_21}<ln(1)[/tex] ; [tex]ln\frac{x_11}{x_21}<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. 



#2
Dec608, 06:38 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

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_11> 0[/itex], [itex]x_2 1> 0[/itex] so [itex]\frac{x_21}{x_11}> 0[/itex] and [itex]ln\frac{x_21}{x_11}[/itex] exists. 



#3
Dec608, 07:53 AM

P: 365

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_1x_2}{x_21})[/tex] [tex]1<\frac{x_1x_2}{x_21}<0[/tex] [tex]0<1+\frac{x_1x_2}{x_21}<1[/tex] and out of here [tex]ln(1+\frac{x_1x_2}{x_21})<ln(1)[/tex] [tex]ln(1+\frac{x_1x_2}{x_21})<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_11}{x_21}[/tex] so that [tex]0<\frac{x_11}{x_21}<1[/tex] and out of here [tex]ln\frac{x_11}{x_21}<0[/tex] 



#4
Dec608, 01:40 PM

P: 365

[prove] monotonicity of function
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. 



#5
Dec608, 09:59 PM

P: 179





#6
Dec708, 12:15 PM

P: 365

Can somebody please help with f(x)=xsin(x), E=(0,п)
x_{1}<x_{2} [tex]f(x_1)f(x_2)=x_1sin(x_1)x_2+sin(x_2)=x_1x_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 ; 1sin(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 ; 1sin(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. 


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