MATH116 Lecture Notes - Lecture 17: Regular Polygon, Opata Language, Maryland Route 2

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Practice solutions 9: use di erentials (or a linearization) to estimate the value of 3. Solution: here it should be clear that we want to use x = 125 as our point of tangency. x we have (for general x) Doing this with the function f (x) = 3 f (x) = f (125) + y. Therefore, when trying to approximate 3 (x 125) 150 = f (150) we have f (150) 5 + 1 + x3 taken over the interval: approximate the area under the curve of the function y = [0, 1], by dividing the interval into 5 equal parts, and making rectangles touching the curve at sample points in the interval. We have x = 1 x4 = 0. 8, and x5 = 1. 5, and xi = 0 + i x = i: that is, x0 = 0, x1 = 0. 2, x2 = 0. 4, x3 = 0. 6, (a) the area using left endpoints is:

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