40_Cal_Solution of Calculus_6e

# 40_Cal_Solution of Calculus_6e - C01S0M.045 Suppose that a...

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Unformatted text preview: C01S0M.045: Suppose that a , b , and c are arbitrary real numbers. Then | a + b + c | = | ( a + b ) + c | 5 | a + b | + | c | 5 | a | + | b | + | c | . . C01S0M.046: Suppose that a and b are arbitrary real numbers. Then | a | = | ( a − b ) + b | 5 | a − b | + | b | . Therefore | a | − | b | 5 | a − b | . C01S0M.047: If x − 3 > 0 and x + 2 > 0, then x > 3 and x > − 2, so x > 3. If x − 3 < 0 and x + 2 < 0, then x < 3 and x < − 2, so x < − 2. Answer: ( −∞ , − 2) ∪ (3 , ∞ ). C01S0M.048: ( x − 1)( x − 2) < 0: x − 1 and x − 2 have opposite signs, so either x < 1 and x > 2 (which leads to no values of x ) or x > 1 and x < 2. Answer: (1 , 2). C01S0M.049: ( x − 4)( x + 2) > 0: Either x > 4 and x > − 2 (so that x > 4) or x < 4 and x < − 2 (so that x < − 2). Answer: ( −∞ , − 2) ∪ (4 , + ∞ ). C01S0M.050: 2 x = 15 − x 2 : x 2 + 2 x − 15 = 0, so ( x − 3)( x + 5) = 0. Now x + 5 > x − 3, so x − 3 = 0 or x + 5 5 0. Thus x = 3 or x 5 − 5. Answer: (5....
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## This note was uploaded on 09/14/2011 for the course MATH 101 taught by Professor Cheng during the Spring '11 term at Ecole Hôtelière de Lausanne.

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