Question

Let f(x)=201+9e−3x .


Over what interval is the growth rate of the function decreasing?


(ln3/9, ∞)


(−∞, ln 3/9)


(ln9/3, ∞)


(−∞, ln 9/3)

Answer 1

Answer: Third option is correct.

Step-by-step explanation:

Since we have given that

$f(x)=\frac{20}{1+9e^{-3x}}$

As we know that it is a logistic function, and there are two types of growth rate :

1) Increasing Growth rate .

2)Diminishing  Growth rate .

Here, Carrying capacity = 20

So, at half of its carrying capacity growth rate changes.i.e.

at y>10, there is diminishing growth rate.

AT y<10, there is increasing growth rate.

so, put y=10

so, it becomes,

$10=\frac{20}{1+9e^{-3x}}\\\\1+9e^{-3x}=\frac{20}{10}=2\\\\9e^{-3x}=2-1\\\\9e^{-3x}=1\\\\e^{-3x}=\frac{1}{9}\\\\\text{Taking logrithms on both sides,}\\\\-3x=\ln 1-\ln9\\\\-3x=-\ln 9\\\\3x=\ln 9\\\\x=\frac{\ln 9}{3}$

So, the interval will be $(\frac{ln9}{3},\infty)$

Above this there will growth rate at a decreasing rate and below this the growth rate at a increasing rate.

Hence, Third option is correct.