(i) Since P(t) gives the population of the culture after t seconds, the population after 1 second is
P(1) = 3•1² + 3e¹ + 10 = 13 + 3e ≈ 21.155
In Mathematica, it's convenient to define a function:
P[t_] := 3t^2 + 3E^t + 10
(E is case-sensitive and must be capitalized. Alternatively, you could use Exp[t]. You can also specify that the argument t must be non-negative by entering a condition via P[t_ ;/ t >= 0], but that's not necessary.)
Then just evaluate P[1], or N[P[1]] or N <at symbol> P[1] or P[1] // N to get a numerical result.
(ii) The average rate of change of P(t) over an interval [a, b} is
(P(b) - P(a))/(b - a)
Then the ARoC between t = 2 and t = 6 is
(P(6) - P(2))/(6 - 2) ≈ 321.030
In M,
(P[6] - P[2])/(6 - 2)
and you can also include N just as before.
(iii) You want the instantaneous rate of change of P when t = 60 (since 1 minute = 60 seconds). Differentiate P :
P'(t) = 6t + 3e^t
Evaluate the derivative at t = 60 :
P'(60) = 6•60 + 3e⁶⁰ = 360 + 3e⁶⁰
The approximate value is quite large, so I'll just leave its exact value.
In M, the quickest way would be P'[60], or you can differentiate and replace (via ReplaceAll or /.) t with 60 as in D[P[t], t] /. t -> 60.
