Answer:
Ok, we have:
d(t) = 3*t^2 + 5*t - 2.
The first interval is:
(2, 3)
and remember that, for an interval (a,b), the difference quotient is:
D = (f(b) - f(a))/(b -a)
Then in this first interval we have:
[tex]d = \frac{d(3) - d(2)}{3-2} = \frac{3*3^2 + 5*3 - 2 - (3*2^2 + 5*2 - 2)}{1} = 20[/tex]
So the average rate of change in (2,3) is 20.
now, in (2, 2.5) we have:
[tex]d = \frac{d(2.5) - d(2)}{3-2} = \frac{3*2.5^2 + 5*2.5 - 2 - (3*2^2 + 5*2 - 2)}{1} = 9.25[/tex]
So here the rate of change is 9.25
And in the interval (2, 2.1) we have:
[tex]d = \frac{d(2.1) - d(2)}{3-2} = \frac{3*2.1^2 + 5*2.1 - 2 - (3*2^2 + 5*2 - 2)}{1} = 1.73[/tex]
So in this interval the rate of change is 1.73
