+0

0
80
3
+483

Let F(x)=12/3+ae^kx with f(0)=2 and f(1)=1/2

The value of a is ____???

The value of k is ____????

F(3)_____????

Mar 25, 2020

#1
+23558
+3

*** deleted ***

I solved    12/3    + aekx

When the question SHOULD have read    12 / ( 3+aekx)           Use the correct syntax people !  It IS VERY IMPORTANT !!!!

Mar 25, 2020
edited by ElectricPavlov  Mar 25, 2020
edited by ElectricPavlov  Mar 26, 2020
#2
+1956
+2

Nice, EP!

CalTheGreat  Mar 25, 2020
#3
+111321
+1

I think this is supposed to be a logistics  growth/decay  equation.....we  have   that

12

f(x)   =           _________

a + e^(kx)

We know that   f(0)  = 2      this makes   e^(kx)   =  e^(k*0)  =  1

So

12

2 =        _____

a + 1

Means that   a  =   5

And   f(1)  =  1/2      ....so.......

12

1/2    =            _________           rearrange as

5 + e^(k)

5 + e^(k)  =  12   /   (1/2)

5  +  e^k   =   24        subtract    5 from both sides

e^k =    19

This   implies  that

Ln e^k  = Ln 19

k * Ln e   =   Ln 19

k * (1)  = Ln 19

k = Ln 19

f(3)   =           12   /  [ 5  + e^(3 * Ln 19) ]   =      1  / 572

Here's the graph : https://www.desmos.com/calculator/9lngaoxfod

Mar 26, 2020