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  
Global Moderator Joined: May 2007 Posts: 6,629 Thanks: 622  Re: I'm Missing something Quote:
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  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: I'm Missing something Quote:
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Quote:
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:
 

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