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Answer:

Given `x=w/(1+e^(-x))+c,w>0,cinRR` find the number of solutions. Let `f(x)=w/(1+e^(-x))+c`

(1) `w/(1+e^(-x))+c` is a logistics function. With `w>0` it is increasing everywhere. Also, the function has two horizontal asymptotes; `lim_(x->oo)f(x)=w+c` and `lim_(x->-oo)=-(w+c)` .

(2) We are asked to find the number of times the logistics function crosses the line `y=x` .

(3) **For any choice of `w` , it is possible to find a `c` so that the functions intersect exactly once.** However, for a set of `w` values there can be two or three intersections. For the functions to intersect more than once, the slope of a point on `f(x)` must be greater than or equal to one.

(4) ` `Computing `d/(dx)[w/(1+e^(-x))+c]` we get `f'(x)=(we^(-x))/((1+e^(-x))^2)` .

(5) If the functions intersect three times, they intersect at (0,0), and two other points symmetric about the origin. So without loss of generality let `c` be such that `f(0)=0` .

(6) We find `f'(0)=w/4` .

(7) If `w<=4` then the functions intersect in exactly one point regardless of your choice of `c` ` `.(The maximum slope at any point is 1. If `w=4` you can choose `c` so that the curves intersect at the origin. Then for `x>0` `f'(x)<1` and `f(x)` is below `y=x` , with a symmetric argument for `x<0` thus they intersect once.` `)

(8) If `w>4` the functions can intersect either 2 or 3 times, depending on your choice of `c` . If we choose `c` so that the functions intersect at the origin, consider the following:

`f'(x)>1` on some interval `[0,b)`and `f(0)=0` so `f(x)>x` on that interval, or `f(x)` is above the line `y=x` . But `f'(x)` approaches 0 as x increases without bound, so `f(x)` eventually crosses `y=x` and then remains below it from then on.

(9) Graph with w=2; c=-3 in red,c=2 in black and c=-1 in blue:

Graph with w=4;c=-4 red,c=-2 black,c=0 blue

Graph with w=10;c=-5 black, c=-3.2 red,c=0 blue

243

Steven Harris

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