Test 2 Review 2.pdf - MATH 1432 EXAM 1 REVIEW Answers 1...

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MATH 1432 EXAM 1 REVIEW: Answers 1. Inverse Functions (a) ( i ) Yes; f ( x ) = 3 x 2 + 4 > 0 for all x = f is one-to-one. ( ii ) No; for example g ( - 2) = g (0) = g (2) = 3; g is not one-to-one. ( iii ) Yes; h ( x ) = - 2 e - 2 x < 0 for all x . (b) ( i ) domain: (0 , ) ( ii ) domain: ( -∞ , ) ( iii ) domain: [ - 1 , 1] range: ( -∞ , ) range: (0 , ) range: - π 2 , π 2 (c) ( i ) f - 1 ( x ) = 3 1 - x x ( ii ) range: all x such that x = - 1. (d) ( f - 1 ) (3) = 1 f (1) = 1 2 / 3 = 3 2 . (e) ( G - 1 ) ( - 1) = 1 G ( π ) = 1 - 1 = - 1 Log and exponential functions (a) ln 2 (b) x = 1 2 e (c) 1 2 (d) y = x + 2 (e) ln 2 - 7 24 (f) -1 1 2 3 x -1 0.4 y (g) ( i ) points of inflection: ( 1 , e - 1 / 2 ) , ( - 1 , e - 1 / 2 ) ( ii ) V = 1 0 2 πx e - x 2 / 2 dx = 2 π 1 - e - 1 / 2 3. Derivatives transcendental functions (a) 1 (1 + x 2 ) tan - 1 x + 2 | 2 x + 1 | (2 x + 1) 2 - 1 (b) g ( x ) = e x 2 x 1 - (ln x ) 2 + 2 x e x 2 sin - 1 (ln x ) (c) h ( x ) = 5 x sinh ( e 2 x ) e 2 x (2) - cosh ( e 2 x ) 5 x ln 5 5 2 x (d) y = (sinh 2 x ) x 3 2 x 3 cosh 2 x sinh 2 x + 3 x 2 ln (sinh 2 x ) 4. Integrals of transcendental functions (a) cosh 2 x sinh 2 x dx = 1 3 (cosh 2 x ) 3 / 2 + C (substitution: u = cosh 2 x ) 1
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(b) e ln( x 2 +1 dx = ( x 2 + 1 ) dx = 1 3 x 3 + x + C (c) [ln( x + 2)] 2 x + 2 dx = 1 3 [ln( x + 2)] 3 + C (substitution: u = ln( x + 2)) (d) sec(ln x ) x dx = ln [sec(ln x ) + tan(ln x )] + C (substitution: u = ln x ) (e) ln 2 0 e x 4 - e 2 x dx = sin - 1 e x 2 ln 2 0 = π 12 (substitution: u = e x ) (f) tan - 1 x 1 + x 2 dx = 1 2 ( tan - 1 x ) 2 + C (substitution u = tan - 1 x ) 5. Exponential growth and decay (a) P ( t ) = 2000 e t 2 ln(3 / 2) = 2000 ( 3 2 ) t/ 2 ( i ) P (8) = 2000 ( 3 2 ) 4 = 10 , 125 (
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