How do you sketch the curve f(x)=e^x/(1+e^x)f(x)=ex1+ex ?

1 Answer
Sep 25, 2014

Information from f(x)f(x)

f(0)=1/{1+1}=1/2 Rightarrowf(0)=11+1=12 y-intercept: 1/212

f(x) > 0 Rightarrowf(x)>0 x-intercept: none

lim_{x to infty}e^x/{1+e^x}=1 Rightarrow H.A.: y=1

lim_{x to -infty}e^x/{1+e^x}=0 Rightarrow H.A.: x=0

So far we have the y-intercept (in blue) and H.A.'s (in green):

enter image source here

Information from f'(x)

f'(x)={e^xcdot(1+e^x)-e^xcdot e^x}/{(1+e^x)^2}=e^x/(1+e^x)^2>0

Rightarrow f is always increasing.

Information from f''(x)

f''(x)={e^x cdot (1+e^x)^2-e^xcdot2(1+e^x)e^x}/{(1+e^x)^4}

={e^x(1+e^x)(1-e^x)}/{(1+e^x)^4}={e^x(1-e^x)}/{(1+e^x)^3}

f''(x)>0 on (-infty,0) and f''(x)<0 on (0, infty)

f is concave upward on (-infty,0) and downward on (0, infty).

Hence, we have the graph of f (in blue):

enter image source here