P(t) = 1 / (1 + e^(-kt)), where k is a real number. Show that d²P/dt² = 0 when P = 1/2?
I found that
dP/dt = -(-k)e^(-kt)/(1 + e^(-kt))²
theen from what I understand d²P/dt² is the d/dt of dP/dt?
When I got the derivative of -(-k)e^(-kt)/(1 + e^(-kt))² by using the quotient rule and everything, i got something REALLY complicated:
-k² ( 1 + e^(-kt) )² + 2k² (1 + e^(-kt))
-------------------------------------------
.......................e ^ (kt)
I got d²P/dt² , but how can I plug in P = 1/2? Where exactly is "P" in the messy stuff above?
Thanks
Update:pls help! 10 pts given promptly!
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Verified answer
dP/dt = -(-k)e^(-kt)/(1 + e^(-kt))² is > 0,
e^(-kt) >0 and so is (1 + e^(-kt))²
P(t) is increasing.
d²P/dt² is 0, there is a pt of inflection.
Set yr answer = 0 since d²P/dt² is 0
-k² ( 1 + e^(-kt) )² + 2k² (1 + e^(-kt))
--------------------------------------…
.......................e ^ (kt)
= 0
-k² ( 1 + e^(-kt) )² + 2k² (1 + e^(-kt))=0
( 1 + e^(-kt) ) = 2
e^(-kt) ) = 1
t = 0
Inflection point (0, 1/2)
Substitute t =0 to show that d²P/dt² = 0
Hope that may help.