ECONOMICS 207 SPRING 2006 LABORATORY EXERCISE 5 KEY. 8 = 10(5x 2) = 9(3x + 8), x 50x 20 = 27x x = 92 x = 4. 8x 2 22x + 15 = 0 (2x 3)(4x 5) = 0
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1 ECONOMICS 07 SPRING 006 LABORATORY EXERCISE 5 KEY Problem. Solve the following equations for x. a 5x 3x + 8 = 9 0 5x 3x = 0(5x ) = 9(3x + 8), x x 0 = 7x + 7 3x = 9 x = 4 b 8x x + 5 = 0 8x x + 5 = 0 (x 3)(4x 5) = 0 x = 3, or x = 5 4 c 0x + 4x 4 = 0 0x + 4x 4 = 0 (4x + 6)(5x 4) = 0 x = 3, or x = 4 5 Date: 8 December 006.
2 ECONOMICS 07 SPRING 006 LABORATORY EXERCISE 5 KEY d 79x 5/4 343x / = 0, (x = ) 79x 5/4 343x / = 0 79x 5/4 = 343x / x / = 79 x 5/4 343 x 3/4 = 79 x = ( = (9 7 )3 ) = ( 9 7) 4 =
3 ECONOMICS 07 SPRING 006 LABORATORY EXERCISE 5 KEY 3 Problem. Solve the following systems of equations for x and x using the method of substitution. a {x = 8, x = 6} 44x / x /4 3 = 0 7x / x 3/4 8 = 0 Take the ratio of these two equations, we have, 44x / x /4 = 3 7x / x 3/4 8 x = 3 x 8 x = 6 8 x Substitute x = 6 8 x into the first equation, we obtain, ( ) /4 44x / 6 8 x = 3 44x / 3 4 x ) /4 = 3 ( ) 44 x /4 = 3 3 x /4 = 3 3 x = ( 4 ( 3 44 = 3 ) 4 = 3 4 = 8 x = 64 8 x = 64 8 = 64 8 So, x = 8, x = 64.
4 4 ECONOMICS 07 SPRING 006 LABORATORY EXERCISE 5 KEY b {x = 9, x = } 5x / x /5 x / x 3/5 5x / x /5 5 = 0 x / x 3/5 36 = 0 Take the ratio of these two equations, we have, = x x = 5 36 x = 9 x Substitute x = 9 x into the first equation, we obtain, ( ) /5 5x / 9 x = 5 x / ( 9 x ) /5 = 3 ( ) /5 x /0 = 9 3 x /0 = 3 ( 3 ) 4/5 = ( 3 )/5 ( ) (/5) 0 x = = ( 3 3 ) = 9 x = 9 x = 9 9 = So, x = 9, x =.
5 ECONOMICS 07 SPRING 006 LABORATORY EXERCISE 5 KEY 5 Problem 3. Solve the following systems of equations for x and x first using the method of substitution and then using the method of elimination. a {x =, x = 3} Method of substitution : x 3x = 5 4x + 3x = 7 By the first equation, we can get, x = (5 + 3x )/ Substitute x = (5 + 3x ) into the second equation, we obtain, 4 (5 + 3x ) + 3x = x + 3x = 7 9x = 7 x = 3 x = (5 + 3x ) (5 9) = = So, the solution is x =, x = 3. Method of elimination : Multiply the first equation by - and add it to the second equation. This gives, (x 3x ) = 5 4x + 6x = 0 4x + 3x = 7 3x = 9 x 3x = 5 4x + 6x + 4x + 3x = 7 9x = 7 x = 3 Multiply the above equation by 3 and add it to the first equation. This gives, x = 4 x =. So, the solution is x =, x = 3.
6 6 ECONOMICS 07 SPRING 006 LABORATORY EXERCISE 5 KEY b {x =, x = 4} x + 4x = 3x + 6x = 30 Method of substitution : By the first equation, we can get, x = ( 4x ) = x 6 Substitute x = x 6 into the second equation, we obtain, 3 (x 6) + 6x = 30 6x 8 + 6x = 30 x = 48 x = 4 x = x 6 = So, the solution is x =, x = 4. Method of elimination : Multiply the first equation by 3 and add it to the second equation. 3x + 6x = 8 3x + 6x = 30 x = 48 x = 4. Multiply the above equation by 6 and add it to the second equation. 6x = 4 3x + 6x = 30 3x = 6 x = So, the solution is x =, x = 4.
7 ECONOMICS 07 SPRING 006 LABORATORY EXERCISE 5 KEY 7 Problem 4. Solve the following system of equations for x, x, and x 3 first using the method of substitution and then using the method of elimination. {x =, x =, x 3 = 3} x + x + 6x 3 = 8 x 3x 0x 3 = 3 x + 3x + x 3 = 34 By the first equation we have, x = 8 x 6x 3 Substitute the above equation into the second equation, we have, (8 x 6x 3 ) 3x 0x 3 = 3 (8 x 3 ) + 3(5 x 3 ) + x 3 = 34 x + x 3 = 5 x = 5 x 3 x = 8 (5 x 3 ) 6x 3 = 8 x 3 Substitute the above two equations into the third equation, we have, 6 4x x 3 + x 3 = 34 x 3 = 3 x = 8 x 3 = 8 6 = x = 5 x 3 = 5 6 = So, the solution is x =, x =, x 3 = 3 Method of elimination: Multiply the first equation by and add it to the second equation. This gives, (x + x + 6x 3 ) = 8 x + 4x + x 3 = 36 x 3x 0x 3 = 3 x + x 3 = 5 Second equation plus the third equation, we get, x 3x 0x 3 = 3 x + 3x + x 3 = 34 x 3 = 3 Multiply the above equation by - and add it to the equation x + x 3 = 5. This gives, x 3 = 6 x + x 3 = 5 x =
8 8 ECONOMICS 07 SPRING 006 LABORATORY EXERCISE 5 KEY Multiply the above equation by -, multiply the equation x 3 = 3 by -6 and add it to the first equation, we obtain x = 6x 3 = 8 x + x + 6x 3 = 8 So, the solution is x =, x =, x 3 = 3 x =
9 ECONOMICS 07 SPRING 006 LABORATORY EXERCISE 5 KEY 9 Problem 5. Find the derivatives of each of the following functions with respect to x. a y = x + 4x 3 d y dx = 4x + x b f(x) = x e x f (x) = 6x + 4 e x c f(x) = 3x 4 log[x] f (x) = 6x 4 x d f(x) = 3x + x 4 x f (x) = 6x + log[4] 4 x e f(x) = 3x / + x /3 4x f (x) = 3 x / + 4x /3 + 8x 3 f f(x) = 4x 3 x e x f (x) = x 4 e x x e x
10 0 ECONOMICS 07 SPRING 006 LABORATORY EXERCISE 5 KEY g f(x) = 9x /3 + x log[x] h f(x) = (x + 5) 3 f (x) = 3x /3 + 4x log[x] + x x = 3x /3 + 4x log[x] + x Find in two different ways. f(x) = (x + 5) 3 = 8x x + 50x + 5 f (x) = 4x + 0x + 50 Also, f (x) = 3(x + 5) = 4x + 0x + 50 i f(x) = 4x x +x f(x) = f (x) = 4x x + x = 4x x + 4(x + ) 4x 8 (x + ) = (x + ) j f(x) = 3x +x 4x +5 f (x) = (6x + )(4x + 5) (8x)(3x + x) (4x + 5) = 8x + 30x + 0 (4x + 5) k f(x) = x 4 + e x f (x) = 8x 3 + e x = 8x 3 + 4e x
11 ECONOMICS 07 SPRING 006 LABORATORY EXERCISE 5 KEY l f(x) = 3x e x x +3 f (x) = (6xex + 3x e x )(x + 3) x(3x e x ) (x + 3) = ex (8x + 3x 4 + 9x ) (x + 3) Problem 6. Find the derivatives of each of the following functions with respect to x. a f(x) = (x + x) f (x) = (x + x)(x + ) = 4x 3 + x + 8x b f(x) = (5x )(3x + 4) Show two ways. f(x) = (5x )(3x + 4) = 5x + 4x 8 f (x) = 30x + 4 Also, f (x) = 5(3x + 4) + 3(5x ) = 30x + 4 c f(x) = 4x e x +x f (x) = 4e x +x + e x +x (x + ) 4x = (4 + 8x + 8x)e x +x d f(x) = x e x +3x f (x) = ln x e x +3x + e x +3x (4x + 3) x = x e x +3x (ln +4x + 3)
12 ECONOMICS 07 SPRING 006 LABORATORY EXERCISE 5 KEY e f(x) = x e x 5x f (x) = xe x 5x + e x 5x (4x 5)x = e x 5x (4x 3 5x + x) f f(x) = log[(x 3 4x) ] f (x) = (x 3 4x) (x3 4x) (3x 4) = 6x 8 (x 3 4x) g f(x) = 3x ex 4x + f (x) = (3 ex + 6x e x )(4x + ) (8x)(3x e x ) (4x + ) = ex (4x 3 x + x + 6) (4x + ) h f(x) = 3x log[x ] x + x f (x) = (3 log[x ] + 3x 4x x )(x + x) (x + )(3x log[x ]) (x + x) = (3 log[x ] + 6)(x + x) (x + )(3x log[x ]) (x + x) Problem 7. For each of the following, take the derivative with respect to x, set the derivative equal to zero and solve the resulting equation for x. a {x = 56} f(x) = 56x 3/8 3x
13 ECONOMICS 07 SPRING 006 LABORATORY EXERCISE 5 KEY 3 df(x) dx = x 5/8 3 = 0 x 5/8 = x = = 8 = 56 3 = 8 56 = 3 8 = 5 b {x = 6} f(x) = 8x /4 4x df(x) dx = 8 4 x 3/4 4 = 0 x 3/4 4 = 8 = 8 = 3 4 x = = 4 = 6 c f(x) = 3px /4 4x df(x) dx = 8px 3/4 4 = 0 x 3/4 = 4 8p = (p) x = (p) = (p) 3 d f(x) = 30x 3/5 x /5 6x 3x f(x) x = 8x /5 x /5 6 = 0 x /5 = 3 x /5 ( ) 5 x = 3 x /5 5 / = 3 x
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![Math 181, Exam 1, Study Guide 2 Problem 1 Solution. =[17ln 5 +3(5)] [17 ln 1 +3(1)] =17ln = 17ln5+12](/thumbs/83/87603917.jpg "Math 181, Exam 1, Study Guide 2 Problem 1 Solution. =[17ln 5 +3(5)] [17 ln 1 +3(1)] =17ln = 17ln5+12")
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