
#1
Mar1410, 05:35 PM

P: 6

1. The problem statement, all variables and given/known data
A scientist predicted that the population of fish in a lake could be modeled by the function f(t)= 40t/(t^2+1), where t is given in days. The function that actually models the fish population is g(t)=45t/(t^2+8t+7). Determine where g(t)>f(t). 2. Relevant equations f(t)= 40t/(t^2+1) g(t)=45t/(t^2+8t+7) g(t)>f(t) 3. The attempt at a solution g(t)>f(t) 45t/(t^2+8t+7)>40t/(t^2+1) 45t/(t+1)(t+7)40t/(t^2+1)>0 Find LCD by multiplying 1 45t/(t+1)(t+7) x (t^2+1)/(t^2+1)40t/(t^2+1) x (t+7)(t+1)/(t+7)(t+1) > 0 Simplifies to (5t^3320t^2235t)/(t+1)(t+7)(t^2+1) 5t(t^264t47)/(t+1)(t+7)(t^2+1) >0 Am i doing this correct? I don't know what to do next. 



#2
Mar1410, 07:26 PM

P: 30

Another way of doing it is to find where f(t) and g(t) intersect and then evaluate the equations at values a little bit off those intersection points to find which one is higher
So i suggest you solve: [tex]\frac{40t}{(t^2+1)}[/tex] = [tex]\frac{45t}{(t^2+8t+7)}[/tex] Step one should be multiplying both sides by [tex](t^2+8t+7)[/tex] and [tex](t^2+1)[/tex] 


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