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.

Explanation:

Since we have given that

`f(x)=(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=(20)/(1+9e^(-3x))\\\\1+9e^(-3x)=(20)/(10)=2\\\\9e^(-3x)=2-1\\\\9e^(-3x)=1\\\\e^(-3x)=(1)/(9)\\\\\text{Taking logrithms on both sides,}\\\\-3x=\ln 1-\ln9\\\\-3x=-\ln 9\\\\3x=\ln 9\\\\x=(\ln 9)/(3)`

So, the interval will be
`((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.