Question

H(t)=(t+3)^2+5

What is the average rate of change of h over the interval −5≤t≤−1?

Answer 1

The average rate of change is 0

Solution:

The average rate of change of function is given as:

`\text{ Rate of change} = (f(b) - f(a))/(b-a)`

Here the given function is:

`h(t) = (t + 3)^2+5`

The interval given is
`-5\leq t\leq -1`

The average rate of change of the function h(t) over the interval
`-5\le t\le -1` can be calculated as:

`\text{ Rate of change } = (h(-1)-h(-5))/((-1)-(-5))`

Find h( -5 ) and h( -1 ):

Substitute t = -5 in given h(t)

`h(-5) = (-5 + 3)^2+5 = 4 + 5 = 9`

Substitute t = -1 in given h(t)

`h(-1) = (-1+3)^2+5\\\\h(-1) = 4 + 5 = 9`

Thus average rate of change is given as:

`\text{ Rate of change } =(9-9)/(-1-(-5)) = 0`

Thus the average rate of change is 0