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College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson

College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson

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## College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson

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**College Algebra**Fifth Edition James StewartLothar RedlinSaleem Watson**Absolute Value of a Number**• Recall from Section P.3 that the absolute value of a number a is given by**Absolute Value of a Number**• It represents the distance from a to the origin on the real number line**Absolute Value of a Number**• More generally, |x – a| is the distance between x and a on the real number line. • The figure illustrates the fact that the distance between 2 and 5 is 3.**E.g. 1—Solving an Absolute Value Equation**• Solve the inequality |2x – 5| = 3 • The equation |2x – 5| = 3 is equivalent to two equations 2x – 5 = 3 or 2x – 5 = –3 2x = 8 or 2x = 2x = 4 or x = 1 • The solutions are 1 and 4.**E.g. 2—Solving an Absolute Value Equation**• Solve the inequality 3|x – 7| + 5 = 14 • First, we isolate the absolute value on one sideof the equal sign. 3|x – 7| + 5 = 14 3|x – 7| = 9 |x – 7| = 3x – 7 = 3 or x – 7 = –3x = 10 or x = 4 • The solutions are 4 and 10.**Properties of Absolute Value Inequalities**• We use these properties to solve inequalities that involve absolute value.**Properties of Absolute Value Inequalities**• These properties can be proved using the definition of absolute value. • For example, to prove Property 1, the inequality |x| < c says that the distance from x to 0 is less than c.**Properties of Absolute Value Inequalities**• From the figure, you can see that this is true if and only if x is between -c and c.**Solution 1**E.g. 3—Absolute Value Inequality • Solve the inequality |x – 5| < 2 • The inequality |x – 5| < 2 is equivalent to –2 < x – 5 < 2 (Property 1) 3 < x < 7 (Add 5) • The solution set is the open interval (3, 7).**Solution 2**E.g. 3—Absolute Value Inequality • Geometrically, the solution set consists of: • All numbers x whose distance from 5 is less than 2.**Solution 2**E.g. 3—Absolute Value Inequality • From the figure, we see that this is the interval (3, 7).**E.g. 4—Solving an Absolute Value Inequality**• Solve the inequality |3x + 2| ≥ 4 • By Property 4, the inequality |3x + 2| ≥ 4 is equivalent to: 3x + 2 ≥ 4 or 3x + 2 ≤ –4 3x≥ 2 3x≤ –6 (Subtract 2) x≥ 2/3 x≤ –2 (Divide by 3)**E.g. 4—Solving an Absolute Value Inequality**• So, the solution set is: {x | x≤ –2 or x ≥ 2/3} = (-∞, -2] [2/3, ∞)**E.g. 5—Piston Tolerances**• The specifications for a car engine indicate that the pistons have diameter 3.8745 in. with a tolerance of 0.0015 in. • This means that the diameters can vary from the indicated specification by as much as 0.0015 in. and still be acceptable.**E.g. 5—Piston Tolerances**• Find an inequality involving absolute values that describes the range of possible diameters for the pistons. • Solve the inequality.**Example (a)**E.g. 5—Piston Tolerances • Let d represent the actual diameter of a piston. • The difference between the actualdiameter (d) and the specified diameter (3.8745) is less than 0.0015. • So, we have |d – 3.8745| ≤ 0.0015**Example (b)**E.g. 5—Piston Tolerances • The inequality is equivalent to • –0.0015 ≤ d – 3.8745 ≤ 0.0015 • 3.8730 ≤ d ≤ 3.8760 • So, acceptable piston diameters may varybetween 3.8730 in. and 3.8760 in.