Explanation Given information: f ( x ) = x / ( x 2 − 1 ) , g ( x ) = ( 1 − x ) / ( 1 + x ) and h ( x ) = 2 / ( 3 x + 1 ) . Explanation: Let, f ( x ) = x / ( x 2 − 1 ) and g ( x ) = ( 1 − x ) / ( 1 + x ) . To determine the expression for f ( x ) − g ( x ) , f ( x ) − g ( x ) = x ( x 2 − 1 ) − ( 1 − x ) ( 1 + x ) , x ≠ − 1 , 1 . The restriction x ≠ − 1 , 1 comes from the fact that f ( x ) is defined only for x ≠ − 1 , 1 and g ( x ) is defined only for x ≠ − 1 . To subtract these two fractions, their denominators must be the same. A common denominator for x ( x 2 − 1 ) and ( 1 − x ) ( 1 + x ) is ( x 2 − 1 ) . The denominator of first fraction is already ( x 2 − 1 ) . Thus, multiply only the second fraction ( 1 − x ) ( 1 + x ) by x − 1 x − 1 , to obtain an equivalent expression whose common denominator is ( x 2 − 1 ) . Thus, x ( x 2 − 1 ) − ( 1 − x ) ( 1 + x ) = x ( x 2 − 1 ) − ( 1 − x ) ( 1 + x ) ⋅ ( x − 1 ) ( x − 1 )

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ISBN: 9780137638826

Author: Goldstein

Publisher: PEARSON

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Chapter 0, Problem 25RE

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The value of f(x)−g(x) for f(x)=x/(x2−1), g(x)=(1−x)/(1+x) and h(x)=2/(3x+1).

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Chapter 0, Problem 24RE

Chapter 0, Problem 26RE

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The value of f ( x ) − g ( x ) for f ( x ) = x / ( x 2 − 1 ) , g ( x ) = ( 1 − x ) / ( 1 + x ) and h ( x ) = 2 / ( 3 x + 1 ) .

Solution Summary: The author explains how to determine the value of f(x) — a common denominator is used to subtract the two tions.

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The value of f(x)−g(x) for f(x)=x/(x2−1), g(x)=(1−x)/(1+x) and h(x)=2/(3x+1).

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