How do you find the average rate of change of #f(t) = 1.1t^2 - 2.5t + 1.5# over the interval [3,5]?

Question

Elijah Abram · Accepted Answer

6.3

The average rate of change of a function over an interval #[a, b]# is the same as the slope of the line going through the points #(a, f(a))# and #(b, f(b))#. As an expression, this is #\frac{f(b)-f(a)}{b-a}#. In this case, the line goes through (3, 3.9) and (5, 16.5). Thus, the average rate of change of #f(x)# over the interval #[3, 5]# is equal to #\frac{16.5-3.9}{5-3}=6.3#

Luke Alvarez · Answer

undefined To find the average rate of change of the function f(t) = 1.1t^2 - 2.5t + 1.5 over the interval [3,5], we use the formula:

Average rate of change = (f(b) - f(a)) / (b - a)

where 'a' and 'b' represent the endpoints of the interval.

1. Substitute t = 3 and t = 5 into the function to find f(3) and f(5).
2. Calculate the difference between f(5) and f(3).
3. Divide the difference by the length of the interval, which is (5 - 3).

Let's calculate it:

1. Find f(3):

f(3) = 1.1(3)^2 - 2.5(3) + 1.5 = 1.1(9) - 7.5 + 1.5 = 9.9 - 7.5 + 1.5 = 3.9

2. Find f(5):

f(5) = 1.1(5)^2 - 2.5(5) + 1.5 = 1.1(25) - 12.5 + 1.5 = 27.5 - 12.5 + 1.5 = 16.5

3. Calculate the difference:

f(5) - f(3) = 16.5 - 3.9 = 12.6

4. Calculate the average rate of change:

Average rate of change = (f(5) - f(3)) / (5 - 3) = 12.6 / 2 = 6.3

So, the average rate of change of f(t) over the interval [3,5] is 6.3.