Question

What is the average rate of change for the function h(x) = –5x2 + 12x over the interval 2 ≤ x ≤ 5?

Answer 1

The average rate of change for the function h(x) = –5x^2 + 12x over the interval 2 ≤ x ≤ 5 is -23.

Step-by-step explanation:

The average rate of change for the function h(x) = –5x^2 + 12x over the interval 2 ≤ x ≤ 5 can be found by calculating the slope of the secant line between the two points.

First, we need to find the values of h(x) at x = 2 and x = 5:

h(2) = -5(2)^2 + 12(2) = -20 + 24 = 4

h(5) = -5(5)^2 + 12(5) = -125 + 60 = -65

Then, we can calculate the average rate of change using the formula:

Average Rate of Change = (h(5) - h(2)) / (5 - 2) = (-65 - 4) / (5 - 2) = -69 / 3 = -23

Therefore, the average rate of change for the function h(x) over the interval 2 ≤ x ≤ 5 is -23.

Answer 2

1. use the function f(x) = -x^2 + 3x + 10 to answer the following.

a. on the interval [2,6], what is the average rate of change?

b. on the interval (2,6) when does the instantaneous rate of change equal the average rate of change?

1. a. -5

b. x = 4

Step-by-step explanation: