By Stefan Waner
Filled with suitable, assorted, and present real-world functions, Stefan Waner and Steven Costenoble's FINITE arithmetic AND utilized CALCULUS, 6th variation is helping you relate to arithmetic. a good number of the purposes are in keeping with genuine, referenced info from enterprise, economics, the existence sciences, and the social sciences. Thorough, sincerely delineated spreadsheet and TI Graphing Calculator guide appears to be like through the booklet. Acclaimed for its clarity and supported by way of the authors' well known web site, this booklet can assist you snatch and comprehend mathematics--whatever your studying type should be. on hand with InfoTrac scholar Collections http://gocengage.com/infotrac.
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Extra info for Finite mathematics and applied calculus
Example text
The kinds of equations that we have in mind are equations in x and y, such as √ y = 4x−1, 2x 2 − y = 0, y = 3x 2 + 1, y = x − 1 . The graph of an equation in the two variables x and y consists of all points (x, y) in the plane whose coordinates are solutions of the equation. EXAMPLE 2 Graph of an Equation Obtain the graph of the equation y − x 2 = 0 . Solution We can solve the equation for y to obtain y = x 2. Solutions can then be obtained by choosing values for x and then computing y by squaring the value of x, as shown in the following table: x −3 −2 −1 0 1 2 3 y = x2 9 4 1 0 1 4 9 Plotting these points (x, y) gives the following picture (left side of Figure 5), suggesting the graph on the right in Figure 5.
X − 1)(x + 2) + (x − 1)(x + 4) = 0 (x 2 − 1)[(x + 2) + (x + 4)] = 0 Factor the left-hand side. (x 2 − 1)(2x + 6) = 0 Either P = 0 or Q = 0. Either x 2 − 1 = 0 or 2x + 6 = 0 x = −3, −1 , or 1. Solve the individual equations. 6 Solving Miscellaneous Equations 31 EXAMPLE 1 Solving by Factoring Solve 12x(x 2 − 4) 5 (x 2 + 2) 6 + 12x(x 2 − 4) 6 (x 2 + 2) 5 = 0 . Solution Again, we start by factoring the left-hand side: 12x(x 2 − 4) 5 (x 2 + 2) 6 + 12x(x 2 − 4) 6 (x 2 + 2) 5 = 12x(x 2 − 4) 5 (x 2 + 2) 5 [(x 2 + 2) + (x 2 − 4)] = 12x(x 2 − 4) 5 (x 2 + 2) 5 (2x 2 − 2) = 24x(x 2 − 4) 5 (x 2 + 2) 5 (x 2 − 1) .
9 9 4 √ √ √ 4 6 43. 45. 9 + 16 44. √ 5 25 √ √ √ √ 47. 9 + 16 48. 25 − 16 46. 25 − 16 √ √ 3 4 50. 81 − 16 51. 3 27/8 49. 8 − 27 √ 3 8 × 64 2 5x −3 x 72. √ x 75. 73. 1 3 2 3x √ 3 x 5 4 − √ + √ 79. 4 3 x 3x x √ 5 2 3 x 7 81. − √ 4 2 x3 1 3 − 3 83. 2 3 (x + 1) 4 (x 2 + 1) 76. 1 3 4x 2 x 3/2 2 + − 2 3 6 3x √ 3 7 5 x + √ 80. √ − 8 5 x 23 x 78. 1 2 82. 8x √x − √ 5 3 x3 84. 2 3 3 (x 2 + 1) 7 − 3(x 2 + 1) −3 4 Convert the expressions in Exercises 85–96 to radical form. 85. 22/3 86. 34/5 89. (x 1/2 y 1/3 ) 1/5 90.