Let f (x) = 15/ 1+4e^-0.2x
What is the point of maximum growth rate for the
logistic function f(x)? Show all work.
Round your answer to the nearest hundredth.
Hint: Your answer should be an ordered pair.

  (6.931, 7.5)

Step-by-step explanation:

The point of maximum growth of a logistic function is always halfway between the horizontal asymptotes. Here, those are y=0 and y=15, so the point of interest is where f(x) = 15/2:

  15/2 = 15/(1+4e^(-.2x)) . . . . . use 15/2 for f(x)

  2 = 1 +4e^(-.2x) . . . . . . . . . . . match denominators

  1/4 = e^(-.2x) . . . . . . . . . . . . . subtract 1, divide by 4

  ln(1/4)/-0.2 = 5·ln(4) = x ≈ 6.93147 . . . . . . take natural logs, evaluate

The ordered pair (x, f(x)) is (6.93147, 7.5).

D. Kelsey graphed the slope as the y-intercept and the y-intercept as the slope.

Step-by-step explanation:

The equation to graph was : y= 3x + 1 where the slope is 3 and the y-intercept is 1

However, Kelsey graph line passes through points (-3, 0)and (0, 3).From these points the equation of the line is;

m=Δy/Δx  where

m=the slope of the graph

Δy= 3-0 =3

Δx= 0--3 =3

m=3/3 = 1

The equation of the line should be;

m=Δy/Δx

1=y-3/x-0

1x= y-3

3+1x= y

y= 1x + 3 ----------where the slope is 1 and y-intercept is 3

So you can see here that in the original equation, the slope m is 3 and y-intercept is 1 as shown in the first attached graph.

While in the Kelsey's graph, the slope is 1 and the y-intercept is 3 as in the second attached graph.

Thus, the correct answer is D : Kelsey graphed the slope,3 as the y-intercept and the y-intercept,1, as the slope.

Answer choice A is incorrect because Kelsey didnot graph the y-intercept, 1 on the x-axis but graphed point (-3,0) on the x-axis.

Answer choice B is incorrect because on Kelsey's equation the y-intercept is 3 and not -3. i.e. y=1x + 3

Answer choice C is incorrect because Kelsey graphed the slope as up 1 right 1 which is okay per the equation , y=1x+3.However, this incorrect because the correct graph has a slope of 3.