Sarahisme Posted May 18, 2005 Share Posted May 18, 2005 lighting quick question.... how do you show this? i dunno how to include abs values and stuff.... Thanks Sarah Link to comment Share on other sites More sharing options...
matt grime Posted May 18, 2005 Share Posted May 18, 2005 Well it's false so I wouldn't try to prove it, unless you're failing to mention what g is. Link to comment Share on other sites More sharing options...
Sarahisme Posted May 18, 2005 Author Share Posted May 18, 2005 oops :S, my mistake, g is g(x) = (x+9)^(1/3) - x , and the interval is [2,3] Link to comment Share on other sites More sharing options...
matt grime Posted May 18, 2005 Share Posted May 18, 2005 So, you know, by the MVT that there is y in the interval [u,v] such that f(v)-f(u) = f'(y)(v-u) whenever v and u are in [2,3] or |f(v)-f(u)| = |f'(y)||v-u| now, what's the minimum value that f'(y) can be if y is some number in [2,3]? Link to comment Share on other sites More sharing options...
Sarahisme Posted May 18, 2005 Author Share Posted May 18, 2005 when y = 3? Link to comment Share on other sites More sharing options...
Sarahisme Posted May 18, 2005 Author Share Posted May 18, 2005 which is a minimum value of..... f'(3) = -0.9364 Link to comment Share on other sites More sharing options...
Sarahisme Posted May 18, 2005 Author Share Posted May 18, 2005 i mean f'(2) is the minimum = -0.9326 Link to comment Share on other sites More sharing options...
matt grime Posted May 18, 2005 Share Posted May 18, 2005 Sorry I meant to say what is the maximal value of |f'| on that interval for then |f(v)-f(u)| = |f'(y)||v-u| < max{|f'|}|v-u| so as long as the max of |f'| is less than a quarter we're ok. Link to comment Share on other sites More sharing options...
Sarahisme Posted May 18, 2005 Author Share Posted May 18, 2005 oh ok, yes that makes sense. thanks matt -Sarah Link to comment Share on other sites More sharing options...
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