March 8th, 2011, 10:57 AM  #1 
Newbie Joined: Mar 2011 Posts: 1 Thanks: 0  Linearization :( HELP!
Using a calculator or computer, generate a graph like the one shown below by graphing y=e^(2x)1 and y=2x for 0.5 < x < 0.5. (What is the relationship between y=e^(2x)1 and y=2x?) So...I'm given a graph with 2 lines here (1 curved and 1 straight  would that be the tangent line or something?) but I can't post the graph. 1. Use your graph to estimate to one decimal place the (largest) magnitude of the error in approximating e^(2x)1 by 2x for 0.5 < x < 0.5. error: 2. Is the approximation an over or an underestimate when x > 0? (Enter over, or under.) THANK YOU SO MUCH IN ADVANCE FOR HELPING ME! 
March 9th, 2011, 09:15 PM  #2 
Member Joined: Jul 2009 Posts: 57 Thanks: 0  Re: Linearization :( HELP!
Hey, I should be able to help you a little. I'm not fully sure what the question is asking for when it says, "(largest) magnitude", but nonetheless  [note, if interval really is 0.5 < x < 0.5 where x cannot be 0.5, you will have to change this up a bit] is the linearization of: at a = 0. Now, let and Then the tangent line, L(x) at a = 0: The "max" amount of error from point where a = 0 would be at either or . Both have the same max error. The max error is found by: So with an error of +/ 0.5 at x = a = 0: Says that with an error of 0.5 in x, there is a max error of 1 in y AT . Now, if by the question "magnitude" does it mean error from at , then you would simply take the absolute value of the difference of and at : I say take it at 0.5 as is greatest at within the interval [0.5, 0.5] So this implies that there is an error of value e at = 0.5 in . Considering that , it is an underestimate. If the interval however is indeed 0.5 < x < 0.5 where x cannot be 0.5, then error would approach e. I'm not sure if I did all this correctly, but hopefully I did a descent job helping you out anyway. 

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