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 February 11th, 2014, 07:21 PM #1 Member   Joined: Feb 2014 Posts: 91 Thanks: 1 I'm Missing something A bicyclist is riding on a path modeled by the function f(x) = 0.04(8x - x^2) where and are measured in miles. Find the rate of change of elevation at x = 2 Here are the steps I've taken before I got stuck. I) f(x) = 0.04(8x - x^2) II) f(x) = 0.32x - 0.04x^2 III) f^-1(2) = lim 0.32(2 + h) - 0.04(2^2)/h IV) f^-1(2) = lim 0.64 + 0.32h - 0.08/h What's my next step from this point. I think I'll be able to figure it out from there.
February 12th, 2014, 04:57 PM   #2
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Re: I'm Missing something

Quote:
 Originally Posted by The_Ys_Guy A bicyclist is riding on a path modeled by the function f(x) = 0.04(8x - x^2) where and are measured in miles. Find the rate of change of elevation at x = 2 Here are the steps I've taken before I got stuck. I) f(x) = 0.04(8x - x^2) II) f(x) = 0.32x - 0.04x^2 III) f^-1(2) = lim 0.32(2 + h) - 0.04(2^2)/h IV) f^-1(2) = lim 0.64 + 0.32h - 0.08/h What's my next step from this point. I think I'll be able to figure it out from there.
I have no idea what steps III and IV are supposed to mean. I assume f(x) is elevation.
If that is correct, then f'(x) = .32 - .08x, so f'(2) = .16 (rate of change of elevation).

February 12th, 2014, 05:35 PM   #3
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Re: I'm Missing something

Quote:
 Originally Posted by The_Ys_Guy A bicyclist is riding on a path modeled by the function f(x) = 0.04(8x - x^2) where and are measured in miles. Find the rate of change of elevation at x = 2
So f(x) is the elevation as a function of distance traveled.

Quote:
 Here are the steps I've taken before I got stuck. I) f(x) = 0.04(8x - x^2) II) f(x) = 0.32x - 0.04x^2
Okay, but I don't think I would have done this. Just use 8x- x^2 and multiply the result by .04

Quote:
 III) f^-1(2) = lim 0.32(2 + h) - 0.04(2^2)/h IV) f^-1(2) = lim 0.64 + 0.32h - 0.08/h
These are wrong. For one thing you don't mean f^-1, which represents the inverse function. You can represent the derivative by f'.
f'(2)= lim [f(2+h)- f(2)]/h. You don't have the "2+h" in the square
f'(2)= lim [{.32(2+ h)- .04(2+h)^2}- {.32(2)- .04(2^29}]/h= lim [{.64- .32h- .04(4+ 4h+ h^2)}- {.64- .16}]/h
Reduce that. Notice that the constant terms, .64- .04(4)- (.64+ .16), cancel so every term involves an "h". Cancel the "h" in the denominator and take the limit as h goes to 0.

Quote:
 What's my next step from this point. I think I'll be able to figure it out from there.

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