Given h(x) = 1/(2 e^x + 2)

Question

1. Determine whether ℎ(x) is increasing or decreasing.

2. What is the range of ℎ(x)?

3. Find an expression for the inverse function ℎ^-1(x).

Answer 1

Graph this!

h(x) is decreasing over the whole real line and the range is (0,1)

For the inverse, interchange the roles of x and y

x=(1/2)[1/(e^y+1)]

2x=1/(e^y+1)

1/(2x)=e^y+1

[1/(2x)]-1= e^y

y=ln{[1/(2x)-1]} and of course x<1/2