M-273Q Midterm: MATH 273 Montana State ReviewE2S17

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turquoisegnu128

15 Feb 2019

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Hint: for e ective study, explain why if true" and give a counterexample if false. " (a) There exists a function f with continuous second-order partial derivatives such that fx(x, y) = x + y2 and fy(x, y) = x y2. If f (x, y) = ln y, then f (x, y) = T or f lim (x,y) (a,b) f (x, y) = l. If f (x, y) l as (x, y) (a, b) along every straight line through (a, b), then (b) (c) (d) (e) If f has a local minimum at (a, b) and f is di erentiable at (a, b), then f (a, b) = 0. 1 x2 + y2 + 1 (a) find equations for the following level curves for f, and sketch them. (a) f (x, y) = 1. S in terms of s and t only: let f (x, y, z) = xzex+y2.

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ceruleangoose957

5. (18 pts) Circle True if the statement is ALWAYS true; circle False otherwise. No explanations are required. Let f be a one-to-one function and f its inverse. (a) If the point (2, 5) lies on the graph off, then the point (5,2) lies on the graph of f. True False f (2) = 5 f-1 (5) = 2 (b) sin-1 (T) = 0 True False Since >1, ris not in the domain of sin-?; | or sin-1 (7) #0, since sin (0) = 0 # Iff and g are two functions defined on (-1, 1), and if (c) limg (x) = 0, then it must be true that lim[ f(x) g (x)] = 0. True False Let f (x) = 1 / x, if x #0, and f (0) = 4 and g(x) = x. Then lim g (x) = lim (x) = 0, but Lim ( f (x) g(x)] = lim [() x] = 18 0. Iffis continuous on (-1, 1), and if f (0) = 10 and lim g (x) = 2, (a) True False then lin (= 5. Because lim : (*) - Limx-of (*) .f(0) = 20 = 5 If f is continuous on [1, 3], and if f(1) = 0 and f(3) = 4, then the equation f(x) = (e) has a solution in (1, 3). True False Since f continuous on [1, 3), f(1) = 0 < < 4 = f (3), by the Intermidiate Value Theorem there exists a cin (1, 3) such that f (c) = . (f) Let f be a positive function with vertical asymptote x = 5. Then lim f (x) = +0. True False Let f (x) = 1, if x < 5, and f (x) ==, ifx>5. Then lim f (x) = +00, and therefore, X = 5 is a vertical asymptote. 5 - X So, f is a positive function with vertical asymptote x = 5 but lim f (x) + +co, since lim f (x) = + 0 + lim f (x) = 1, which means that lim f (x) does not exist.

M-273Q Midterm: MATH 273 Montana State ReviewE2S17

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