By S. Waner, S. Costenoble

ISBN-10: 0495384283

ISBN-13: 9780495384281

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**Additional resources for Applied Calculus [enh rvw edn.]**

**Sample text**

82/3 = ( 8) 2 = 22 = 4 √ √ 31/2 3 1 1 1 1 6 1/2−1/3 1/6 −3/2 = = 3 = 3 = 3 = 3/2 = √ = 3 = 4. √ 3. 9 3 1/3 3 3 9 3 27 3 ( 9) √ 5. 22 27/2 = 22 23+1/2 = 22 23 21/2 = 25 21/2 = 25 2 ■ Example 4 Simplifying Algebraic Expressions Simplify the following. a. (x 3 ) 5/3 x3 b. √ 4 a6 c. (x y) −3 y −3/2 √ x −2 y Solution x5 (x 3 ) 5/3 = = x2 x3 x3 √ √ 4 b. a 6 = a 6/4 = a 3/2 = a · a 1/2 = a a a. c. x −3 y −3 y −3/2 1 1 (x y) −3 y −3/2 = = −2+3 1/2+3+3/2 = 5 √ −2 −2 1/2 x y x y x y xy Converting Between Rational, Radical, and Exponential Form In calculus we must often convert algebraic expressions involving powers of x, such 3 3 as 2 , into expressions in which x does not appear in the denominator, such as x −2 .

Example 3 Factoring Quadratics Factor the following: a. 4x 2 − 5x − 6 b. x 4 − 5x 2 + 6 Solution a. Possible factorizations of 4x 2 are (2x)(2x) or (x)(4x). Possible factorizations of −6 are (1)(−6), (2)(−3). We now systematically try out all the possibilities until we come up with the correct one. (2x)(2x) and (1)(−6): (2x + 1)(2x − 6) = 4x 2 − 10x − 6 (2x)(2x) and (2)(−3): (2x + 2)(2x − 3) = 4x 2 − 2x − 6 (x)(4x) and (1)(−6): (x + 1)(4x − 6) = 4x 2 − 2x − 6 (x)(4x) and (2)(−3): (x + 2)(4x − 3) = 4x 2 + 5x − 6 Change signs: (x − 2)(4x + 3) = 4x 2 − 5x − 6 No good No good No good Almost!

X 2 − x + 1 = 0 2 2 23. x 2 − x = 1 24. 16x 2 = −24x − 9 1 25. x = 2 − x 1 26. x + 4 = x −2 2 2 2 2 37. y 3 + 3y 2 + 3y + 2 = 0 39. x 3 − x 2 − 5x + 5 = 0 40. x 3 − x 2 − 3x + 3 = 0 41. 2x 6 − x 4 − 2x 2 + 1 = 0 42. 3x 6 − x 4 − 12x 2 + 4 = 0 43. (x 2 + 3x + 2)(x 2 − 5x + 6) = 0 44. 6 Solving Miscellaneous Equations Equations that are not polynomial equations of low degree often arise in calculus. Many of these complicated-looking equations can be solved easily if you remember the following, which we used in the previous section: Solving an Equation of the Form P ؒ Q = 0 If a product is equal to 0, then at least one of the factors must be 0.

### Applied Calculus [enh rvw edn.] by S. Waner, S. Costenoble

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