Math 141: PRACTICE TEST # 1: Chapter 1 KEY



Math 141 Practice Test # 1- KEY: Chapter 1, 2.1, 6.1 – 6.3

|1. |Not a function. (Input of 4 has 2 outputs.) D = {4, 8} R = {2, 3, 5} |

|2. |Function (Substituting values for x yield only one answer.) D = {x| x > - 3} or [ - 3, [pic]) |

| | R = {y| y > 0} or [0, [pic]) |

|3. |Function (Passes vertical line test) D = (-[pic],[pic]) R = [- 4, [pic]) |

|4. |Not a function (Input of 1 has 2 outputs.) D = {1, 3, 5} R = {2, 4, 6} |

|5. |D |

|11. |A. |[- 7, 4] |B. |[- 2, 4] |C. |(- 4.5, 0), (2, 0) |D. |(0, 3.5) |

|12. |[pic] |a. Graph x |- 4 |-3 |-2 |0 | |

| | | y |- 1 |0 |1 |- 3 | |

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| | |b. D = ( - [pic], [pic] ) |

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| | |c. R = (-[pic], 1] |

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| | |d. x-intercepts: (-3, 0), (-1.5, 0); y –intercept: ((0, - 3) |

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|13. |[pic] |a. |

| | |b. D = [ - 2, [pic]) |

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| | |c. R = ( - [pic], 5] |

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| | |d. x-intercept (2, 0); y-intercept (0, 3) |

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|14. |a. |Odd |Ck for even |Ck for odd |

| | | |f(x) = f(-x) |f(-x) = - f(x) |

| | | |x = - x NO |-x = - (x) |

| | | | |-x = - x YES |

| | | | | |

| |b. |Even |Ck for even |No need to check for Odd |

| | | |f(x) = f(-x) | |

| | | |-2x2 = -2(-x)2 | |

| | | |-2x2 = -2x2 YES | |

| | | | | |

| |c. |Neither |Ck for even |Ck for odd |

| | | |f(x) = f(-x) |f(-x) = - f(x) |

| | | |x3 – 1 = (-x)3 – 1 |(-x)3 – 1 = - (x3 – 1) |

| | | |x3 -1 = - x3 – 1 NO |- x3 – 1 = - x3 + 1 NO |

| | | | | |

| |d. |Neither |Ck for even |Ck for odd |

| | | |f(x) = f(-x) |f(-x) = - f(x) |

| | | |[pic] NO |[pic] NO |

| | |

|15. |a. |(f∙g)(x) = (3x + 1)(- 4x – 3) → - 12x2 – 9x – 4x – 3 → - 12x2 – 13x|16. |[pic] |

| | |– 3 | | |

| | | | | |

| |b. |(g-f)(x) = (- 4x – 3) – (3x + 1) → - 4x – 3 – 3x – 1 → - 7x - 4 | | |

| | | | | |

| |c. |[pic] = [pic] | | |

| | | | | |

| |d. |(f + g)(x) = (3x + 1) + (- 4x – 3) → 3x + 1 – 4x – 3 → - x - 2 | | |

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| | |Answer |Work / thinking |

|17. |A. |m = - 2 |Coefficient of x |

| |B. |(0, 3) |Let x = 0, find y. y = -2(0) + 3 ( y = 3 |

| |C. |Decreasing |Negative slope |

| |D. |Zero at 3/2 or (3/2, 0) |Let y = 0, find x. 0 = -2x + 3 ( -3 = -2x ( 3/2 = x |

| |E. |-2 or -2 / 1 |Average rate of change is the same as slope for linear functions. |

|17. |F. |[pic] |18. F. |[pic] |

|18. |A. |m = - 1 / 3 |Coefficient of x |

| |B. |(0, 1) |Let x = 0, find y. |

| |C. |Decreasing |Negative slope |

| |D. |Zero at 3 or (3, 0) |Let y = 0, find x. 0 = -1/3x + 1 ( -1 = -1/3x ( (Multiply by -3/1) ( 3 = x |

| |E. |-1 / 3 |Average rate of change is the same as slope for linear functions. |

|19. | |5 / 7 |Avg. rate = (y / (x = (y2 – y1) / (x2 – x1) = (7-2)/(4- -3) = 5 / 7 |

|20. |A. |Linear |Constant rate of change: as x values increase by 1, y values increased by 2 |

| |B. |Linear |Constant rate of change: as x values increase by 1, y values increased by 0 |

| |C. |Not linear |Rate of change is NOT constant: as x values increase by 1, y values increase by 3, then 4, 5, and 6 |

|21. | |(3, 3) |Solve the 1st equation for y by subtracting 2x from both sides. ( y = -2x + 9. Substitute for y in the 2nd |

| | | |equation. ( 3x – 2(-2x + 9) = 3 ( Use the distributive property 3x + 4x – 18 = 3. Combine like terms 7x – 18 =|

| | | |3 ( Add 18 to both sides of the equation 7x = 21 ( Divide by 7 to get x = 3. Substitute x = 3 into the 1st |

| | | |equation ( 2(3) + y = 9 ( Subtract 6 from both sides of equation to get y = 3. |

|22. | |(1, -2) |[pic]Adding the two equations gives 43x = 43 or x = 1. Substitute x = 1 into the 1st equation ( 4(1) – 3y = |

| | | |10 ( -3y = 6 ( y = -2 |

|23. | |(5, 1) |[pic]R1 = r1 + r2 ([pic]( R1 = r1 [pic]2 ([pic]( |

| | | |R2 = r1 – r2 ([pic]( x = 5, y = 1 |

|24. |A. |(-2, -12) |[pic]Adding the two equations gives -23x = 46 |

| | | |Or x = - 2. Substitute -2 for x in 1st equation ( 4(-2) – 3y = 28 ( -3y = 36 ( y = -12 |

| |B. |No solution |[pic]Adding the two equations gives 0 = - 22 |

| | | |This is a false number statement ( inconsistent system ( no solution |

| |C. |[pic] |[pic]Adding the two equations gives 0 = 0 |

| | | |This is a true number statement ( dependent system (same line) |

| |D. |(5, 21) |Multiply 1st equation by 6 (the LCD) and the 2nd equation by 2 (the LCD) to get rid of the fractions. New |

| | | |system has the equations 9x - 2y = 3 and 4x – y = - 1. Multiply 4x – y = - 1 by – 2 and add to 9x – 2y = 3 to |

| | | |eliminate the y’s and get x = 5. Substitute x = 5 into 4x – y = - 1 ( 4(5) – y = - 1 ( 20 – y = - 1 ( 21 = y |

|25. |A. |Independent system |X = 1, Y = 2, Z = 3 (unique solution) |

| |B. |Consistent system |x + 2y + 8z = 0, y + 3z = 2, 0 = 0 (dependent solution) Solve for x and y in terms of z ( x = -2y – 8z; y = - |

| | | |3z + 2; z can be any real number |

| |C. |Inconsistent system |No solution because last row 0 = 3 is a false statement. |

|26. | |x = -1, y = 4, z = 0 |Subtract equation 2 from equation 1 (x + y + 6z = 3) – (x + y + 3z = 3) ( 3z = 0 ( z = 0. Using the 1st and |

| | | |3rd equations and substituting z = 0 ( x + y = 3 and x + 2y = 7. Subtracting the (new) 1st equation from the |

| | | |(new) 2nd equation, (x + 2y = 7) – (x + y = 3) ( y = 4. Substituting y = 4 and z = 0 into the original 2nd |

| | | |equation ( x + 4 + 3(0) = 3 ( x = - 1. |

|27. | |1500 children tickets, |Let x = # of children tickets; y = # of adult tickets ( x + y = 2200 (number of tickets) and 1.5x + 4y = 5050 |

| | |700 adult tickets |(money). Multiply the 1st equation by -1.5 and add the two equations ( 2.5y = 1750 ( y = 700. Substitute 1500|

| | | |for y in the original 1st equation ( x + 700 = 2200 ( x = 1500 |

|28. | |Dimensions: 15 cm by 12 cm. |Let x = length, y = width. Area: xy = 180, perimeter: 2x + 2y = 54. Solve for x or y in area equation ( x = |

| | | |180 / y. Substitute into the perimeter equation ( 2(180/y) + 2y = 54. Multiply everything by the LCD = y ( 360|

| | | |+ 2y2 = 54y. Quadratic equation = 0, reduce the equation by dividing by 2, and use quadratic formula ( 2y2 – |

| | | |54y + 360 = 0 ( y2 – 27y + 180 = 0. y = 12 or 15. Substituting into the perimeter equation gives x = 15 or 12.|

|29. | |2.5 lbs of Kenyan, |Let x = lbs of Kenyan, y = lbs. of Sri Lanka. Pound equation x + y = 3; cost equation 3.5x + 5.6 y = 11.55. |

| | |0.5 lbs of Sri Lanka |Multiply pound equation by – 35 (-35x – 35y = -105) and price equation by 10 (35x + 56y = 115.5). Adding the |

| | | |two new equations ( 21y = 10.5 ( y = 0.5. Substituting into the pound equation ( x + 0.5 = 3 ( x = 2.5 |

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