SOLUTION: A rocket was launched from a platform. The function h(t)=16t-5t^2+3 expresses the height in feet of the rocket over time in seconds. Find the average rate of change over the interv

Question 1066494: A rocket was launched from a platform. The function h(t)=16t-5t^2+3 expresses the height in feet of the rocket over time in seconds. Find the average rate of change over the interval (0,2). State the domain and range.
I am having problems finding the Domain.
Average rate of change is: 15-3/2-0 = 6ft/sec
Range is: x=-16/2(-5) = 1.6
x=16(1.6)-5(103)^2+3
R={0,15.5}
Please help me find the Domain.
Thanks

Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
Average rate of change for a function f over an interval [a, b] is
:
(f(b) - f(a)) / (b - a)
:
We are given the following function h and the interval (0, 2)
:
Note average rate of change is usually stated for CLOSED intervals
:
h(t) = -5t^2 +16t +3
:
h(0) = -5(0)^2 +16(0) +3 = 3
h(2) = -5(2)^2 +16(2) +3 = 15
:
Average rate of change = (15 - 3) / (2 - 0) = 6 feet per second
:
****************
Domain is (0, 2)
****************
:
We need to find coordinates for the Vertex of h(t)
:
t = -16 / -10 = 8/5 = 1.6
h(1.6) = -5(1.6)^2 +16(1.6) +3 = 15.8
:
This tells us that the max value(15.8) for h(t) occurs at t = 1.6
:
******************
Range is (3, 15.8]
******************
:
h(t) is a parabola that curves downward, here is its graph
:

:

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