monotonicity without using differentiation

[b!tcoin]

I want to prove the monotonicity of the following function WITHOUT using derivatives.
f(x)=x²-4x+3, on the interval [-1,1]
I tried this:
Let x1<x2, numbers of the interval, where x1<x2
x1²<x2²
add -4x to both sides.
x1²-4x1>x2²-4x2
ADD 3 : x1²-4x1+3>x2²-4x2+3
f(x1)>f(x2)
so because we started with x1<x2 and ended with f(x1)>f(x2) the function is strictly decreasing.
Any ideas whether this is correct, and if there are any other ways of solving this without differentiating.

 

[JeffM]

Your general idea of choosing an arbitrary pair of points in the interval such that

[MATH]-\ 1 \le x_1 < x_2 \le 1[/MATH] is exactly on target.

However, the proposition that [MATH]\therefore x_1^2 < x_2^2[/MATH] is false.

[MATH]-\ 0.5 < -\ 0.25 \text { but } 0.25 = (-\ 0.5)^2 > 0.0625 = (-\ 0.25)^2.[/MATH]
Refine your argument by adding the case where

[MATH]x_1^2 \ge x_2^2[/MATH].

 

[Dr.Peterson]

I'd start by completing the square.

One reason for doing this is the realization that f will be monotonic on either side of the vertex, and a form that emphasizes the vertex will make this visible.

 

[Steven G]

I want to prove the monotonicity of the following function WITHOUT using derivatives.
f(x)=x²-4x+3, on the interval [-1,1]
I tried this:
Let x1<x2, numbers of the interval, where x1<x2
x1²<x2²
add -4x to both sides.
x1²-4x1>x2²-4x2
ADD 3 : x1²-4x1+3>x2²-4x2+3
f(x1)>f(x2)
so because we started with x1<x2 and ended with f(x1)>f(x2) the function is strictly decreasing.
Any ideas whether this is correct, and if there are any other ways of solving this without differentiating.

It seems that you are simply making it work. When you added -4x to both sides did you make sure that -4x1<-4x2 on the interval [-1,1]? How about x1^2<x2^2? You just can't write what you want, it must be true.
For example if x1<0 and x2>0 (so x1<x2 as requested) then -4x1>0 and -4x2<0. Then -4x1<-4x2 is NOT true. So (assuming x1^2<x2^2) you can't jump from x1^2<x2^2 to x1^2-4x1<x2^2-4x2.
You need to use real rigorous math.
Also, as already pointed out, x1^2 <x2^2 is not necessarily true in your interval.

Most importantly, don't you see that the way YOU did your work that you could have concluded that f(x1)<f(x2)????

 

[JeffM]

It seems that you are simply making it work. When you added -4x to both sides did you make sure that -4x1<-4x2 on the interval [-1,1]? How about x1^2<x2^2? You just can't write what you want, it must be true.
For example if x1<0 and x2>0 (so x1<x2 as requested) then -4x1>0 and -4x2<0. Then -4x1<-4x2 is NOT true. So (assuming x1^2<x2^2) you can't jump from x1^2<x2^2 to x1^2-4x1<x2^2-4x2.
You need to use real rigorous math.
Also, as already pointed out, x1^2 <x2^2 is not necessarily true in your interval.

Most importantly, don't you see that the way YOU did your work that you could have concluded that f(x1)<f(x2)????

Thanks jomo. I did not look beyond the issue of the squares, but of course the same issue comes up in multiplying by - 4.