SOLUTION: Hi all, Im having trouble understanding what this question is asking, can anyone show me how to solve? The function V(t) = 11e^(-t)ln(t+1) models the speed of a particle for time

Question 205735: Hi all, Im having trouble understanding what this question is asking, can anyone show me how to solve?
The function V(t) = 11e^(-t)ln(t+1) models the speed of a particle for time
t >=0.
a) Say breifly how it is known whether V(t) grows or decays as t goes to infinity, without calculating or graphing. Explain why (justify answer).
b) Find the derivative of V(t) and factorise.
Help with explanations and steps would be very helpful.
Thanks, -nick.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
The function V(t) = 11e^(-t)ln(t+1) models the speed of a particle for time
t >=0.
a) Say breifly how it is known whether V(t) grows or decays as t goes to infinity, without calculating or graphing. Explain why (justify answer).
e^(-t) causes it to approach zero as t approaches infinity.
---------------
b) Find the derivative of V(t) and factorise.
V'(t) = 11*(-e^(-t)ln(t+1) + e^(-t)/(t+1))
V'(t) = (11e^(-t))*(-ln(t+1) + 1/(t+1))

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