Answer:
(a)Increasing
(b)t=1.34 years
(c)16 cameras per year
Step-by-step explanation:
Given the function
N(t) = 5.85t³-23.43t²+45.06t+69.5, 0≤t≤4
(a)N(0)=5.85(0)³-23.43(0)²+45.06(0)+69.5=69.5
N(4)=5.85(4)³-23.43(4)²+45.06(4)+69.5=249.26
A function is increasing whenever x₁≤x₂, f(x₁)≤f(x₂).
Since in the interval (0,4), N(0)<N(4), we say the function is increasing.
(b)The number of communities using surveillance cameras at intersections changed least rapidly at the point where the derivative of the function is zero.
N(t) = 5.85t³-23.43t²+45.06t+69.5
N'(t)=17.49t²-46.86t+45.06
If N'(t)=0,
17.49t²-46.86t+45.06=0
Solving the quadratic equation gives the values of t as:
t=1.3396-0.8842i
t=1.3396+0.8842i
We take the Real Part as our Minimum value,
The time when number of communities using surveillance cameras at intersections changed least rapidly is:
t=1.34(to 2 decimal places)
(c)Rate of Increase using a security camera/year.
N'(t)=17.49t²-46.86t+45.06
N'(t)=17.49(1)²-46.86(1)+45.06
=15.69
≈16 cameras/year
