MATH 154 Lecture Notes - Lecture 27: Implicit Function, Maxima And Minima

• School: Simon Fraser University
• Department: Mathematics
• Course: MATH 154
• Professor: Yusuf Tuncer

Document Summary

1 + 1 + 1 = 3 so the point is on the curve. Note: don"t forget the brackets or you"ll lose marks! With the curve (cid:1876)(cid:2870)+(cid:1876)(cid:1877)(cid:2871)+(cid:1877)(cid:2870)=(cid:885), show that point (1, 1) is on the curve. Show newton"s method for (cid:1877)=(cid:1876)(cid:2871) (cid:885)(cid:1876)(cid:2870)+(cid:884)(cid:1876) (cid:883) with the initial guess x = 2. On the exam, we don"t have to show every single step. For this question this is all we have to do. If we have a straight wire 24cm long & bend it in 2 places to form 3 sides of a rectangle, express the area as a function of x. Usually polynomials have a domain of all real numbers, but we have a constraint. 2x + y = 24, so y = 24 2x. (cid:4666)(cid:1876)(cid:4667)=(cid:884)(cid:886) (cid:886)(cid:1876), so the only critical point is x = 6. We have to evaluate the function at the endpoints and the critical point.