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MAT133Y1 Study Guide - Final Guide: Lagrange Multiplier, Maxima And Minima

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turquoisegnu128
1 Oct 2018
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UTSG
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Mathematics
Course
MAT133Y1
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All

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MAT133Y1 Full Course Notes
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MAT133Y1 Full Course Notes
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Multiple choice (x2 + e2x)3e(cid:0)2x (1 + x (cid:0) x2)2=3 , then f 0(0) = Let f (x) =(cid:26) e(cid:0)x x (cid:20) 0 ax + b x > 0. If f is di(cid:11)erentiable at x = 0 , then f (x) = 0 when x = where a and b are constants. The graph of y = (1 (cid:0) ln x)2 has. B: an in(cid:13)ection point at x = 1. C: an in(cid:13)ection point at x = 2. D: an in(cid:13)ection point at x = pe. E: an in(cid:13)ection point at x = e2. The largest value attained by y = x ln x if 0 < x (cid:20) e is. E: unde(cid:12)ned et lnjtj dt , then f 00(1) = 1 + ex and the x -axis from x = 0 to x = ln 5 is. The area bounded by the graphs of y = x.

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4. (16 pts) MULTIPLE CHOICE! CIRCLE THE CORRECT ANSWER IN EACH PART. (I) ( 4 pts) Complete the statement of the Intermediate Value Theorem: Suppose f is continuous on the interval [a, b] and L is a number between f (a) and f (b). Then there is at least one number c in (a, b) such that (a) C = L; (b) f(c) = 0; (c) f(c) = L; (a) f' (c) = L; (e) C = 0; (f) f (a) < c < f (b); (g) NONE OF THE PREVIOUS ANSWERS. (II) (4 pts) The figure shows the graph of a function f. At what point (points) c does the conclusion of the Intermediate Value Theorem hold for f (x) on the interval [1, 5] and L = 2. --- - - - -1 - - - 2 - - I- - - (a) c = 1; (b) C = 0; (c) C = 2; (a) c = 3; (e) C = 4; (f) C = 5; (g) c= -1. (III) (4 pts) Given that cos x s f (x) s et, for all x > 0 evaluate lim f (x) x+0+ (a) lim f (x) = 0; (b) lim f (x) = 1; (c) lim f (x) = e; x0+ X0+ (d) lim f(x) DOES NOT EXIST. WE DON'T HAVE ENOUGH INFORMATION TO ANSWER THE QUESTION *40+ (IV) (4 pts) Given the function 1 f (x) = , find its inverse, f+(x) x -2 (a) f (x) = x -2; (b) f«+ (x) = x+2 ; (c) fº+ (x) = -2; (f) f-+ (x) -2; (e) f (x) DOES NOT EXIST.