course Mth 173 ???????????assignment #029s?r??????????????Calculus I
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14:35:41 Query 4.7.24 (was problem 7 p 290 ) prove if g' < h' on (a,b} and g(b) = h(b) then h < g on (a,b)--g,h both cont on [a,b] diff on (a,b)Explain why you expect, that for the given conditions, the function h will be strictly less than the function g on the interval.
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RESPONSE --> Not Problem 4.7.24 4th edition. To figure out what the graph of the two functions look like (and what the values are), we can start at either end. In this case, we don't have any values for the beginning of the interval (a), but we do have values for the end (b). We know that both functions are equal at the end. Now we can consider what the function looks like in the interval itself. We know that the slope of h is greater than the slope of g. However, we know that they must meet at the end of the interval. Therefore the only way this can work is if h starts out the interval at a lower position than g. This means that during the interval, the value of h will be less than the value of g up until the end.
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14:36:17 Since f ' (x) < 0 on the interval the function is decreasing on the interval, hence since f(b) = 0 it follows that f(x) > 0 on the interval. From this it follows that g(x) - h(x) > 0 on the interval and g(x) > h(x). **
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RESPONSE --> Well, that was certainly more concise!
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14:36:51 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> A good recap of l'Hopital's rule, which was certainly needed.
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14:39:28 I was surprised (but not disappointed) that the query was only on one question. I did gain insight in that after I first typed in my original answer, I realized that it was wrong. I had proved (quite successfully, I thought) that the Racetrack principle was wrong! I'm hoping that my revised answer is more correct.
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RESPONSE --> Hmm. I never even considered the Racetrack principle. I see how it applies. And yes, thanks for the short query.
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