A First Course in Calculus, 3rd Edition by Serge Lang PDF By Serge Lang

ISBN-10: 0201042231

ISBN-13: 9780201042238

This 5th variation of Lang's booklet covers all of the issues routinely taught within the first-year calculus series. Divided into 5 elements, each one element of a primary path IN CALCULUS comprises examples and purposes in terms of the subject lined. additionally, the rear of the publication comprises specific recommendations to quite a few the routines, letting them be used as worked-out examples -- one of many major advancements over past variations.

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Additional info for A First Course in Calculus, 3rd Edition

Sample text

27 3 22. -12 26. Λ/5 30. 78 34. 38. 30 10 's2->3 Chapter 1 Summary and Review Examples We will use the margins in the chapter summaries to give examples that correspond to the topic being reviewed whenever it is appropriate. The number(s) in brackets next to each heading indicates the section(s) in which that topic is discussed. 1] a —b a Φ b a < b a fCb a > b a^ b 1. 3 + 5(3 + 4) = 3 + 5(7) = 3 + 35 = 38 a a # a a a is is is is is is equal to b not equal to b less than b not less than b greater than b not greater than b The inequality symbols always point to the smaller quantity.

Here are some further examples of how the addition property of equality can be used to solve equations. ▼ Examples Solve the following equations. 4. x - 5 = 12 SOLUTION Since we want x alone on the left side, we choose to add +5 to both sides. x—5 + 5 = 12+5 x + 0 = 17 JC = 17 Addition property of equality 52 Chapter 2 Linear Equations and Inequalities 5. a + 7 = - 3 SOLUTION Since we want a to be by itself, we will add the opposite of 7 to both sides. a + 7 + (-7) = - 3 + (-7) ö + 0 =-10 a = -l0 Addition property of equality 6.

We choose to multiply both sides by I since (h)(5) = 1. Here is the solution: 5x = (ϊ)(5χ) = (5· 5)x = be = x= 30 (δ)(30) (δ)(30) 6 6 Multiplication property of equality Associative property of multiplication We chose to multiply by 5 because it is the reciprocal of 5. We can see that multiplication by any number but zero will not change the solution set. If, however, we were to multiply both sides by zero, the result would always be 0 = 0, since multiplication by zero always results in zero. The statement 0 = 0 is nice, but we have lost our variable and cannot solve the equation.