let f(x) = exp(-1/x) = 1/exp(1/x)

f(-infinity) = 1 (horizontal asymptote)

increasing and concave up from f(-infinity) to f(0)

f(0-) = +infinity (vertical asymptote when approaching from left)

f(0+) = 0 (one of the things that makes this function curious is that its left- and right-sided limits at 0 differ in existence. what happens on the complex plane?)

right of 0, the function is strictly positive.

f'(x) = exp(-1/x)/x^2

f'(0+) = 0 (slope is horizontal)

all higher derivatives at 0, limit also approaching from right, are 0, making this a "flat function". it's therefore surprising (consider the Taylor series) that the function manages to leave 0.

right of 0, the function is strictly increasing.

f''(x) = exp(-1/x) * (1-2*x) / x^4

concave up until f(1/2) = exp(-2) ~= 0.1353 . this inflection point seemingly comes out of nowhere: exp(x) is positive, increasing, and concave up everywhere, and -1/x is negative, increasing, and concave down in the neighborhood 0<x, both rather boring. but compose them and you get an inflection point. similar inflection points occur in the bell curve exp(-x^2) and log(x)^e.

concave down beyond that (but still increasing).

f'''(x) = exp(-1/x) * (6*x^2 - 6*x + 1) / x^6

third derivative is zero at f((3-sqrt(3))/6 ~= 0.211) ~= 0.00881 and f((3+sqrt(3))/6 ~= 0.789) ~= 0.281 , representing points of maximum concavity (not curvature).

further derivatives are a family of oscillating functions f(x)*polynomial(x)/x^(2*n) , with the degree of the polynomial increasing by one for each derivative.

f(+infinity) = 1 (horizontal asymptote). note that the function also approaches 1 (albeit from above) when x tends to negative infinity. is it 1 at all infinite points on the complex plane (the point at infinity)?

right of x=0, it's kind of a sigmoid.

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