5_Cal_Solution of Calculus_6e

# 5_Cal_Solution of Calculus_6e - C01S01.026: Given g ( t ) =...

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Unformatted text preview: C01S01.026: Given g ( t ) = 3 √ t + 4 = ( t + 4) 1 / 3 , we note that t + 4 is defined for every real number t and the cube root of t + 4 is defined for every possible resulting value of t + 4. Therefore the domain of g is the set R of all real numbers. C01S01.027: Given f ( t ) = √ 1 − 2 t , we observe that 1 − 2 t is defined for every real number t , but that its square root is defined only when 1 − 2 t is nonnegative. We solve the inequality 1 − 2 t = 0 to find that f ( t ) is defined exactly when t 5 1 2 . Hence the domain of f is the interval ( −∞ , 1 2 ¤ . C01S01.028: Given g ( x ) = 1 ( x + 2) 2 , we see that ( x + 2) 2 is defined for every real number x , but that g ( x ), its reciprocal, will be defined only when ( x + 2) 2 6 = 0; that is, when x + 2 6 = 0. So the domain of g consists of those real numbers x 6 = − 2. C01S01.029: Given f ( x ) = 2 3 − x , we see that 3 − x is defined for all real values of x , but that f ( x ), double its reciprocal, is defined only when...
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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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