6_Cal_Solution of Calculus_6e

# 6_Cal_Solution of Calculus_6e - we require that x 6 = 1 so...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: we require that x 6 = 1 so that the fraction is defined. In addition we require that the fraction be nonnegative so that its square root will be defined. These conditions imply that both numerator and denominator be positive or that both be negative; moreover, the numerator may also be zero. But if the denominator is positive then the [larger] numerator will be positive as well; if the numerator is nonpositive then the [smaller] denominator will be negative. So the domain of f consists of those real numbers for which either x − 1 > or x +1 5 0; that is, either x > 1 or x 5 − 1. So the domain of f is the union of the two intervals ( −∞ , − 1] and (1 , + ∞ ). Alternatively, it consists of those real numbers x not in the interval ( − 1 , 1]. C01S01.035: Given: g ( t ) = t | t | . This fraction will be defined whenever its denominator is nonzero, thus for all real numbers t 6 = 0. So the domain of g consists of the nonzero real numbers; that is, the union of the two intervals (...
View Full Document

## 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.

Ask a homework question - tutors are online