My Math Forum Linearization :( HELP!

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 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] $y= 2x$ is the linearization of: $y= e^{(2x)}-1$ at a = 0. Now, let $f(x)= y = e^{(2x)}-1$ and $\frac{d}{dx}f(x)= 2e^{(2x)}$ Then the tangent line, L(x) at a = 0: $L(x)= y = f(a) + f'(a)(x - a) = f(0) + xf#39;(0) = 0 + 2xe^{(2*0)} = 2x$ The "max" amount of error from point where a = 0 would be at either $dx= \Delta x = 0.5$ or $dx= \Delta x = -0.5$. Both have the same max error. The max error is found by: $dy= 2e^{(2x)} * dx$ So with an error of +/- 0.5 at x = a = 0: $dy = 2e^{(2*0)} * 0.5 = 1 dy = 2e^{(2*0)} * (-0.5) = -1$ Says that with an error of 0.5 in x, there is a max error of 1 in y AT $x= 0$. Now, if by the question "magnitude" does it mean error from $y= e^{(2x)}-1$ at $dx= 0 + 0.5$, then you would simply take the absolute value of the difference of $dy$ and $\Delta y$ at $dx= 0.5$: $\Delta y = [e^{(2*0.5)}-1] - [e^{(2*0)}-1] = e - 1 dy = 2e^{(2*0)} * (-0.5) = -1 m_e = |dy - \Delta y| = |-1 - (e - 1)| = |e| = e$ I say take it at 0.5 as $|dy - \Delta y|$ is greatest at $dx= 0.5$ within the interval [-0.5, 0.5] So this implies that there is an error of value e at $dx$ = 0.5 in $y= e^{(2x)}-1$. Considering that $dy < \Delta y$, 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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using a calculator or computer, generate a graph like the one shown below by graphing y = sin x and y = x for

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