how do you differentiate N=100t(e^-1/12)+500 and make equal to zero?
is it e^(-t/12) or just e^(-1/12 ?
e^1/12
dN/dt is simply 100e^(-1/12). a constant function. it is never equal to zero. always equal to the constant
but i need to equate to zero to find the max value
if the initial N occus January 1986, how do i find when the max N will occur?
you may have miswrote the function.
Population growth is e^(at)
Po*e^(rate*t)
yes, the function is N=100te^-1/12+500
e is to the power of -1/12, isn't that right? how do i differentiate it and find the max?
oh, wait, so sorry, it is -t/12, it was written wrong in my notes.
hehe. in that case you must use product rule. (f(x)*g(x))' = f'(x)g(x) + f(x)g'(x)
f(x) = t and g(x) = e^(-t/12). the rest are all constants
so its 100 ( 1*e^(-t/12) + t*(-1/12)*e^(-1/12)
= 100 ( 1 - t/12) (e^(-t/12)) this function is only = 0 when t = 12 (since e^x is never = 0)
i don't understand how you got from the first line of working to (1-t/12), i understand the second bracket
i don't understand what you don't understand :P lol. i factored out e^(-t/12)
oh, okay.
i don't understand how you got t=12
if you make the brackets equal 0, doesn't the first equal zero, but the second -1?
1-t/12=0 so t=-1?????
how do you make the brackets equal zero ( 1 - t/12) -->doesn't this one equal -1 if you divide by 12 (e^(-t/12))-->and this one 0 i'm so confused?
oops i meant times by 12
(1 - t/12) = 0 --> 1 = t/12 ---> t = 12
e^(-t/12) can never = 0 since its the exponential function.
doesn't e^0 equal 1 though, so wouldn't making t=0 make e^0/12-->1
it would. but nobody wants to do that. were looking for t values which make the derivative of the function = 0
okay, i don't really get the last bit, but i get the main gist
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