The figure below shows the graph of $y=x$ together with the graph of
$$f(x)=x+2x^2.$$
Some folks would call zero a neutral fixed point. Does this make sense?
If $$f(0) = 0$$
Then f has a fixed point at 0, shown by$$f(0) = 0 + 2(0)^2$$
Which does in fact equal 0
I think that if we iterate to the right of 0 then it would repulse infinitely up.
If we iterate to the left it would repulse infinitely down.
Lastly it does make sense to call zero a neutral fixed point as that is the point where the lines settle and are tangent.
I am a little confused as to what the neutral means, and I also want to see if I'm correct in saying these graphs repulse.
Find a Neutral Fixed Point.
$$f(x)=x+2x^2$$
$$and$$
$$f(x)=x$$
Algebraically
$$f(0)=(0)+2(0)^2=0$$
$$f(0)=(0)=0$$
Right hand:
As the x values increase the y values become farther to the fixed point (repulsive).
Left hand:
As the x values increase the y values become closer to the fixed point(attractive).
As we get closer to the fixed point from the right hand limit, the iterate is repulsive.
As we get closer to the fixed point from the left hand limit, the iterate is attractive.
I believe that the neutral fixed point is (0,0). Because it is the point is the same on both graphs.
@Chief_Keith By definition, a fixed point $x_0$ is called neutral if $|f'(x_0)|=1$. Now, we haven't really talked in any depth about this thing called the derivative and denoted by $f'$ but, for this function it turns out that $f'(x)=1+4x$ so that $f'(0)=1$ and zero is a neutral fixed point.
As far as the dynamics goes - anything could happen. Points close by might converge to the point or move away from the point or (as this example shows) they might move towards it from one side and away from the other.