Answer:
(ln(3)/1.3, ∞)
Step-by-step explanation:
The growth rate of the function is given by the first derivative:
f'(x) = -93.6e^(-1.3x)/(1 +3e^(-1.3x))^2
The interval on which this is decreasing can be found by looking at the second derivative:
f''(x) = -93.6(1.3e^(-1.3x)(1+3e^(-1.3x))^-2 -e^(-1.3x)(-2)(3)(-1.3)e^(-1.3x)(1+3e^(-1.3x))^-3)
This will be zero when ...
-3e^(-1.3x) +1 = 0 . . . . . . the only numerator factor that can be zero
e^(-1.3x) = 1/3 . . . . . . . . . add 3e^(-1.3x), divide by 3
-1.3x = ln(1/3) = -ln(3) . . take the logarithm
x = ln(3)/1.3 ≈ 0.845086 . . . . . . . divide by the coefficient of x
The function f(x) is decreasing for all values of x larger than this.
