View Full Version : Average Rate of Change
prism941
Sep 29, 2009, 07:18 AM
what formula would I use for R(x)=400+x^2? Size increases from 20 to 30.
Unknown008
Sep 29, 2009, 07:31 AM
I'm sorry, I don't understand your question.
What increases from 20 to 30?
What is r(x) = 400 + x^2?
What are you looking for?
Is r(x) an area, with its function being 400+x^2, and that area increases from 20 to 30?
It would be best for you to post the whole question, and say what you did already, and what you don't understand.
prism941
Sep 29, 2009, 07:41 AM
the revenue function for the annual super bowl party is defined as follows: R(x)=400+x^2; where x is the increase in group size beyond 10 people. What is the average rate of change of revenue if the group size increases from 20 to 30.
Unknown008
Sep 29, 2009, 07:50 AM
Ok, so 'x' is increasing.
Change in 'x' is 10 (from 30 to 20). This is denoted by \frac{dx}{dt}
Now, you have R(x) = 400 + x^2.
Using Chain rule, you can find the rate of change of revenue, which is denoted by \frac{dR}{dt}
\frac{dR}{dt} = \frac{dR}{dx} \times \frac{dx}{dt}
So, find dR/dx. You have the value of x, which is 20 (you take the initial value of x) You can find the value of dR/dx.
Then, you have the value of dx/dt.
Multiply both values together to get the rate of change of your revenue!
i hope this helped! :)
Unknown008
Oct 1, 2009, 09:25 AM
Please, do not start another thread to ask how to differentiate. You could have asked that here, or on the threads that you already posted before. I assumed that you knew how to differentiate, but you should have said that you have difficulty in finding the derivative of a function!
y=400+x^2
Ok, differentiating y, we get:
\frac{dy}{dx} = 2x
When you differentiate, the derivative of ax^n is : anx^{(n-1)}
So, x^2 becomes 2x.
Multiply the coefficient by the power, then decrease the power by one.
x^3 becomes 3x^2
x^4 becomes 4x^3
Now, with a coefficient:
2x^2 becomes (2)(2)x = 4x
2x^3 becomes (2)(3)x^2 = 6x^2
3x^4 becomes (3)(4)x^3 = 12x^3
Is it OK? :)