Question

Sketch the graphs of the following functions:

y = `1/(1 + "e"^-x)`

Answer 1

The Domain of the function f(x) is the entire real line.

ie., `(-oo, oo)`

⇒ `-oo < x < oo` and the range is (0, 1) ie., 0 < f(x) < 1

No ‘x’ intercept for f(x) and when x = 0

y = `1/2`

∴ The ‘y’ intercept is `(0, 1/2)`

f'(x) = `"e"^-x/(1 + "e"^-x)^2`

f'(x) = 0

⇒ which is absurd.

Hence there is no extremum.

No vertical asymptote for the curve exist and the Horizontal asymptotes are y = 1 and y = 0.