NEED ANSWERED ASAP Derivatives of exponential functions

Question

a rumour spreads in such a way that t hours after the rumour starts, the percent of people involved in passing it on is given by P(t)=100(e^-t- e^-4t). what is the highest percentage of people involved in spreading the rumour within the first 3h and when does this occur?

 

- im having difficulties finding the first derivative and the values after i equate the first derivative to zero 
Pls help! 

 

(im also given that the answer is 47.2% at 0.462 hours)

Answer 1

P'(t) = 100(-e^-t + 4e^-4t)

 

P'(t) = 0 when 4e^-4t  = e^-t

 

                            4e^-3t = 1

 

                                      e^-3t = 0.25

 

                                         -3t = ln(0.25)

 

                                             t = -ln(0.25)/3 ≈ 0.462 hours

 

(_____+_____l____-______) ^{sign of P'(t)}

^{0             0.462                   3}

 

Maximum when t = 0.462 hours

 

Maximum value of P = P(0.462) = 100(e^-0.462 - e^-1.848) = 47.2