The maxima of f(x) occur at its critical points, where f '(x) is zero or undefined. We're given f '(x) is continuous, so we only care about the first case. Looking at the plot, we see that f '(x) = 0 when x = -4, x = 0, and x = 5.
Notice that f '(x) ≥ 0 for all x in the interval [0, 5]. This means f(x) is strictly increasing, and so the absolute maximum of f(x) over [0, 5] occurs at x = 5.
By the fundamental theorem of calculus,

The definite integral corresponds to the area of a trapezoid with height 2 and "bases" of length 5 and 2, so


Answer:
The sequence is arithmetic
we know that
In an arithmetic sequence, the difference between consecutive terms is always the same and is called common difference
In this problem we have
3,9,15,21,...
Let
a1=3, a2=9,a3=15,a4=21
a4-a3=21-15=6
a3-a2=15-9=6
a2-a1=9-3=6
The sequence is arithmetic
The common difference is equal to
<u>Answer:</u> 5 years!
<u>Reasoning:</u>
Year 1: 2700 x .70=1890
Year 2: 1890 x .70= 1323
Year 3: 1323 x .70= 926.1
Year 4: 926.1 x .70= 648.27
Year 5: 648.27 x .70=453.79