Each year, the number of cars in Cory's collection (c) increases by a factor of 1.2. After t years, it will be multiplied by that factor "t" times, so the number of cars will be ...
... c(t) = 15·1.2^t
Each year, the number of cars in Roger's collection (r) increases by 1. After t years, it will have increased by t.
... r(t) = 40 +t
<h3>Part B</h3>c(6) = 15·1.2^6 ≈ 44.79 ≈ 45 . . . Cory's cars after 6 years
r(6) = 40 +6 = 46 . . . Roger's cars after 6 years
<h3>Part C</h3>The number of cars is very nearly the same after 6 years. A short time into the 7th year, the number of cars will be the same. A graph shows the number of cars will be the same after about 6 years and 2 months.
There are no algebraic methods to solve the equation
... r(t) = c(t)
... 45 +t = 15·1.2^t
It can be solved by trial and error, or graphically, or by other iterative means. The solution is not an integer, but non-integer numbers of cars don't make sense. It isn't exactly clear what a solution would look like.
year 6: Cory ≈ 45 cars; Roger = 46 cars
year 7: Cory ≈ 54 cars; Roger = 47 cars
Between year 6 and year 7 (closer to the beginning of year 7), Cory will have 46 cars, about the same number Roger has at that time.
Answer:
Minimum 8 at x=0, Maximum value: 24 at x=4
Step-by-step explanation:
Retrieving data from the original question:
1) Calculating the first derivative
2) Now, let's work to find the critical points
Set this
0, belongs to the interval. Plug it in the original function
3) Making a table x, f(x) then compare
x| f(x)
-1 | f(-1)=9
0 | f(0)=8 Minimum
4 | f(4)=24 Maximum
4) The absolute maximum value is 24 at x=4 and the absolute minimum value is 8 at x=0.
