Math

University of Guyana Faculty of Techno put downy EMT 111 - Final Examination - Solutions declination 20, 2010 Directions: Answer only questions. Show all work. Calculators ARE allowed. 1. Evaluate.(9 points) (a) log64 4 . Since 43 = 64 ) 64 1 3 = 4 ) log64 4 = 1 3 (b) log p 10. present log p 10 = log 101=2 = 1 2 log 10 = 1 2 (c) (2x3+x1)(x2+1). exploitation long division we fit 2x with x1 as a remainder. 2. process for x.(9 points) (a) log(x + 2) + log(x 1) = 1 ) (x + 2)(x 1) = 10 ) x2 + x 12 = 0 ) (x + 4)(x 3) = 0 ) x = 3 (b) x2 x  6 ) x2 x 6  0 ) (x 3)(x + 2)  0 ) x 2 [2; 3] (c) 10x+4 = 100 ) 10x+4 = 102 ) x + 4 = 2 ) x = 2 3. Compute the derivative of the following functions.(20 points) (a) s(t) = 1 t 1 t2 p1 t ) s0(t) = 1 t2 + 2 t3 1 2 p t3 (b) h(x) = x54x2 x3 = x2 4 x ) h0(x) = 2x + 4 x2 (c) f(x) = esec x ) f0(x) = esec x sec x tan x (d) g(x) = ln( romaine lettuce x) ) g0(x) = prate x romaine x = t an x (e) s(t) = sin( p 1 t2) ) s0(t) = cos( p 1 t2) ( p 1 t2)0 | {z } 2t 2 p 1t2 ) s0(t) = t cos( p p 1t2) 1t2 1 4. Evaluate the following integrals.

(24 points) (a) Z x p x2 + 2 dx = 1 2 Z p x2 + 2 d(x2 + 2) = 1 2 (x2 + 2)3=2
2 3 + C = 1 3 (x2 + 2)3=2 + C (b) Z 1 0 xex2 dx = 1 2 Z 1 0 ex2 d(x2) = 1 2 (e1 e0) = 1 2 (e 1) (c) Z cos(1 x) 3 dx = sin(1x) 3 + C (d) Z x3 p 4 x2 dx Let x = 2 sin  ) dx = 2 cos d ) Z x3 p 4 x2 dx = Z 8 sin3 
2 cos d 2 cos  = 8 Z sin3 d = 8 Z (1 cos2 ) sin d = 8[cos  + cos3  3 ] + C = 8[ p 4x2 2 + (4x2)3=2 24 ] + C (e! ) Z 5x 2 (x 1)(x 2) dx = Z 3dx x 1 + Z 8dx x 2 = 3 ln jx 1j + 8 ln jx 2j + C (f) Z  4 0 cos3 x sin x dx = Z  4 0 cos3 x d(cos x) =  cos4 x 4   4 0 = (p1 2 )4 4 + 1 4 = 1 16 + 1 4 = 3 16 5. Use the bodyguard that 12 uid ounces is approximately 6:89 cubic inches to nd the dimensions of the 12-ounce can that can be constructed using the...If you want to feel a full essay, order it on our website: BestEssayCheap.com

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