# Applied Calculus, Enhanced Review 4th Edition (with by Stefan Waner

By Stefan Waner

Take calculus into the genuine global with utilized CALCULUS. Authors Waner and Costenoble make utilized calculus effortless to appreciate and proper for your pursuits. And, this textbook interfaces together with your graphing calculator and your place spreadsheet application. Plus it comes with AppliedCalculusNOW. After an easy pre-test, the AppliedCalculusNOW on-line studying method customizes all of the workouts and sophistication details round your own wishes. This version additionally comes with own show with SMARTHINKING, which provides you entry to one-on-one, on-line tutoring support with knowledgeable within the topic. And it offers a digital examine workforce, too-interact with the teach and different scholars utilizing two-way audio, an interactive whiteboard for discussing the matter, and fast messaging.

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Extra info for Applied Calculus, Enhanced Review 4th Edition (with CengageNOW, Personal Tutor Printed Access Card)

Example text

Quick Examples 1. 2(x − 3) is not equal to 2x − 3 but is equal to 2x − 2(3) = 2x − 6. 2. 3. 4. 5. 6. x(x + 1) = x 2 + x 2x(3x − 4) = 6x 2 − 8x (x − 4)x 2 = x 3 − 4x 2 (x + 2)(x + 3) = (x + 2)x + (x + 2)3 = (x 2 + 2x) + (3x + 6) = x 2 + 5x + 6 (x + 2)(x − 3) = (x + 2)x − (x + 2)3 = (x 2 + 2x) − (3x + 6) = x 2 − x − 6 There is a quicker way of expanding expressions like the last two, called the “FOIL” method (First, Outer, Inner, Last). Consider, for instance, the expression (x + 1)(x − 2). The FOIL method says: Take the product of the ﬁrst terms: x · x = x 2 , the product of the outer terms: x · (−2) = −2x , the product of the inner terms: 1 · x = x , and the product of the last terms: 1 · (−2) = −2, and then add them all up, getting x 2 − 2x + x − 2 = x 2 − x − 2.

2x − 3) 2 26. 10x(x 2 + 1) 4 (x 3 + 1) 5 + 15x 2 (x 2 + 1) 5 (x 3 + 1) 4 √ √ 27. (x 3 + 1) x + 1 − (x 3 + 1) 2 x + 1 √ 28. (x 2 + 1) x + 1 − (x + 1) 3 29. 10. (3x + 1) 2 1 2 11. x + x 13. (2x − 3)(2x + 3) 21 1 y− y 12. (x + 1) 3 + (x + 1) 5 30. (x 2 + 1) 3 (x + 1) 4 − 2 3 (x + 1) 7 In Exercises 31–48, (a) factor the given expression; (b) set the expression equal to zero and solve for the unknown (x in the odd-numbered exercises and y in the even-numbered exercises). 14. (4 + 2x)(4 − 2x) 1 1 y+ 15.

However, f (−2) is not deﬁned, since −2 is not in the domain (−2, 10]. If we substitute 0 for x in the formula for f (x), we get f (0) = (0) 2 − 25(0) + 15 = 15 so f (0) = 15. Similarly, f (10) = (10) 2 − 25(10) + 15 = 100 − 250 + 15 = −135 f (−1) = (−1) 2 − 25(−1) + 15 = 1 + 25 + 15 = 41 f (a) = a 2 − 25a + 15 Substitute a for x. f (x + h) = (x + h) 2 − 25(x + h) + 15 Substitute (x + h) for x. = x 2 + 2xh + h 2 − 25x − 25h + 15 Note how we placed parentheses around the number at which we are evaluating the function.