Advertisements
Advertisements
प्रश्न
Sketch the graphs of the following functions:
y = `1/(1 + "e"^-x)`
आलेख
उत्तर
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.
shaalaa.com
Sketching of Curves
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Applications of Differential Calculus - Exercise 7.9 [पृष्ठ ५३]
