college Calculus problem

Mathematics
Tutor: None Selected Time limit: 1 Day

i am stuck on the following problem. here is a pic of the problem

http://imgur.com/MyJHzXa

please include all work/steps in the answers

thanks in advance

Jan 26th, 2015
p(x) = 9.41 -0.19x + 0.09x2

dp/dx is the derivative of p with respect to x (w.r.t)

This is the average rate of change of price x years after 1990

dp/dx = d/dx(9.41 - 0.19x + 0.09x2)
     =d/dx(9.41) + d/dx(-0.19x) + d/dx(.09x2)
     = 0 + (-0.19) + 2 * 0.09 * x
     = -0.19 + 0.18x

ans a) is dp/dx = -0.19 + 0.18x

Now that you know dp/dx ,you can now calculate the average rate of change of price for any year after 1990



Ans b) We need to find the price in 2010 that is 20 years after 1990.So our 'x' value is 20 .
     Plug in 20 for x in p(x)
     p(20) = 9.41 -0.19(20) + 0.09(400)
           = 9.41 - 3.8 + 36
           = 45.41 - 3.8
           = 41.61
Please check the values again with your calculator

Ans c) Now plug in x = 20 on dp/dx 
         (dp/dx)when x = 20 =  -0.19 + 0.18(20)
                            = -0.19 + 3.6
 
                           = 3.41

Jan 26th, 2015

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Jan 26th, 2015
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