Question 1137193
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Part A
r(x) = 9x^2+7x+81 = projected revenue
c(x) = -2x^2-10x-14 = projected cost


p(x) = profit
p(x) = revenue - cost
p(x) = ( r(x) ) - ( c(x) )
p(x) = ( 9x^2+7x+81 ) - ( -2x^2-10x-14  )
p(x) =  9x^2+7x+81  + 2x^2+10x+14  
p(x) =  (9x^2+2x^2)+(7x+10x)+(81+14)
p(x) =  11x^2+17x+95


The profit function is <font size=4 color=red>p(x) =  11x^2+17x+95</font>


x = number of years into the future 
p(x) = profit, in millions of dollars
Example: x = 2 leads to p(x) = 173, telling us that 2 years into the future the profit will be estimated at 173 million dollars.
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Part B


We will use the formula
{{{R = (p(b) - p(a))/(b-a)}}}
where, 
R = average rate of change on the interval a < x < b
p(x) = the profit function
a,b = start and endpoint of the interval


In this case, a = 3 and b = 5. So we'll first need to compute p(3) and p(5)
plug in x = 3
p(x) =  11x^2+17x+95
p(3) =  11(3)^2+17(3)+95
p(3) =  11(9)+17(3)+95
p(3) =  99+51+95
p(3) =  245
then repeat for x = 5
p(x) =  11x^2+17x+95
p(5) =  11(5)^2+17(5)+95
p(5) =  11(25)+17(5)+95
p(5) =  275+85+95
p(5) =  455


Now we can use the average rate of change formula mentioned earlier to get...
{{{R = (p(b) - p(a))/(b-a)}}}


{{{R = (p(5) - p(3))/(5-3)}}} Plug in a = 3 and b = 5


{{{R = (455 - 245)/(5-3)}}} Replace p(5) with 455; replace p(3) with 245


{{{R = 210/2}}}


{{{R = 105}}}


The average rate of change in profit, from year 3 to year 5, is <font size=4 color=red>105 million dollars per year</font>


This represents the average speed in growth rate of the profit over this time span. 


Recall that y = p(x) is measured in millions of dollars. When we subtracted p(b)-p(a), we found the change in millions of dollars. When we computed b-a in the denominator, we found the change in years. Dividing change in profit over change in time yields the units "millions of dollars per year"; or you can think of it as "dollars per year" in a sense. 
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