Water coming out from a fountain is modeled by the function f(x) = -x^2 + 5x + 4 where f(x) represents the height, in feet, of the water from the fountain at different times x, in seconds. What does the average rate of change of f(x) from x = 3 to x = 5 represent?
The water falls down with an average speed of 3 feet per second from 3 seconds to 5 seconds. The water falls down with an average speed of 5 feet per second from 3 seconds to 5 seconds. The water travels an average distance of 3 feet from 3 seconds to 5 seconds. The water travels an average distance of 5 feet from 3 seconds to 5 seconds.
HI!!
it is the average speed you are computing
Averate rate of change has the formula kinda like the slope ;) \[ \text{ Average Rate of Change }=\frac{f(5)-f(3)}{5-3}\]
to find it you need to compute the numbers
what @Mimi_x3 wrote
ooh no dear
\[f(5)-f(3)\neq f(2)\] you gotta compute!
Umm slight mistake there f(5)= -(5)^2 + 5(5) + 4 f(3) = -(3)^2 + 5(3) + 4 You need to plug these in
no shortcuts just compute \[f(5)\] then compute \[f(3)\] then subtract
\[ f(x) = -x^2 + 5x + 4\\ f(5) = -5^2 + 5\times 5 + 4\]
\[f(x) = -x^2 + 5x + 4\\ f(3) = -3^2 + 5\times 3 + 4\]
\[\text{ Average Rate of Change }=\frac{f(5)-f(3)}{5-3}\] \[\text{ Average Rate of Change }=\frac{(-5^2 + 5( 5) + 4)-(-4^2 + 5(4) + 4)}{5-3}\]
f(5) = 4 f(3) = 10 4-10 = -6 So... \[\frac{ -6 }{ 2 }\] ???
ayee whoops I did 4 instead of 3 in my post above -.- Anyways so you are right ... what -6/2???
I have no clue what I'm doing. o.o
hmmm want me to explain????
well anyways what -6/2
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