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Mathematics, 22.06.2019 21:00 ellycleland16

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)

Answers

ansver
Answer from: Savagecabbahg2925

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.


Let f(x)=201+9e−3x . over what interval is the growth rate of the function decreasing?  (ln3/9, ∞) (
ansver
Answer from: HahaHELPP

f(x)=20/1+9e^-3x

C. (ln9/3, ∞)

ansver
Answer from: mettababeeeee
1. f(x)=8/1+3e^(-0.7x)
to get f(-3) we substitute the value and solve the equation as follows:
f(-3)=8/1+3e^(-0.7*3)
f(-3)=8.36737
which can also be written as:
f(-3)-8.36737=0

2. f(x)=20/1+9e^(3x)
to get the y-intercept we let x=0, hence we shall have:
f(0)=20/1+9e^(3*0)
f(0)=20+9*1
f(0)=20+9
f(0)=29
the y-intercept is at point (0,29)

3. f(x)=24/1+3e^(-1.3x)
Given that the expression above has no denominator, then the domain of the expression is all real numbers then the expression has not vertical asymptote.The same applies for horizontal asymptote and the oblique. 

4. f(x)=15/1+4e^(-0.2x)
 The graph is not increasing at any point, this means that it does not have a point of maximum growth.

5. f(x)=24/1+3e^(-1.3x)
The graph of the function is a decay graph, this means that there is no point in which the graph is increasing hence there's not growth rate.
ansver
Answer from: SiegeHatake4534

\bold{D.\ \bigg(-\infty, \dfrac{ln\ 4}{0.2}\bigg)}

Step-by-step explanation:

y=\dfrac{c}{1+ae^{-rx}}\qquad \text{where growth rate increases from}\ -\infty\ to\ \dfrac{ln\ a}{r}

f(x) = \dfrac{15}{1+4e^{-0.2x}}\qquad \rightarrow \qquad a=4, r=0.2

\text{So, the growth rate increases over the interval}\ \bigg(-\infty, \dfrac{ln\ 4}{0.2}\bigg)

ansver
Answer from: JuanTorres7

\bold{D.\ \bigg(-\infty, \dfrac{ln\ 4}{0.2}\bigg)}

Step-by-step explanation:

y=\dfrac{c}{1+ae^{-rx}}\qquad \text{where growth rate increases from}\ -\infty\ to\ \dfrac{ln\ a}{r}

f(x) = \dfrac{15}{1+4e^{-0.2x}}\qquad \rightarrow \qquad a=4, r=0.2

\text{So, the growth rate increases over the interval}\ \bigg(-\infty, \dfrac{ln\ 4}{0.2}\bigg)

ansver
Answer from: Lia97

the answer is:

(ln9/3, infinity)

I just took the quiz


Step-by-step explanation:

ansver
Answer from: brionesoswaldo09

  (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.


Let f(x)=24/1+3e−1.3x .over what interval is the growth rate of the function decreasing? (−∞, ln 31.
ansver
Answer from: azaz1819

1.) 0.31

2.)(0.73,10)

3.)y=0 and y=24

4.)(6.93,7.5)

5.)(in3/1.3,∞)

ansver
Answer from: QueenB2311

Hi just took the test and heres what I got.


Let f(x)=24/1+3e−1.3x .over what interval is the growth rate of the function decreasing? (−∞, ln 31.
ansver
Answer from: jshhs
From -infinity to 0.04 (ln3/9)

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