Let the function f and g be defined as f(x) = x/ x − 1 and g(x) = 2 /x +1 . Compute the sum (f + g)(x) and the quotient (f/g)(x) in simplest form and describe their domains.
(f + g )(x) =
Domain of (f+g)(x):
(f/g)(x) =
Domain of (f/g)(x):
Given f(x) = x / (x-1) and g(x) = 2 / (x+1)
Then f+g(x) = f(x) + g(x)
Thus, f+g(x) = [ x/(x-1) ] + [ 2/(x+1) ]
f+g(x) = [ x(x+1) + 2(x-1) ] / (x-1)(x+1)
f+g(x) = [ x2 + x +2x - 1 ] / (x2 - 12)
f+g(x) = ( x2 +3x -1 ) / (x2 -1)
Hence, f+g(x) = [ x2 +3x -1 ] / [ x2 -1 ]
f+g(x) is not defined when the denominator is zero
Thus, (x+1)(x-1) ≠ 0 implies x ≠ 1 , -1
Domain of f+g(x) is (-
, -1 ) U ( -1 , 1 ) U ( 1 ,
)
f/g (x) = f(x) / g(x)
Thus, f/g(x) = [ x / (x-1) ] / [2 / (x+1) ]
= [ x / (x-1) ] × [ (x+1) / 2 ]
= x(x+1) / 2(x-1)
= ( x2 + x ) / (2x -2)
Hence f/g (x) = [ x2 +x ] / [ 2x -2 ]
f/g (x) is not defined when the denominator is zero
Thus, 2x-2 ≠ 0
Thus, 2x ≠ 2 implies x ≠1
Domain is ( -
, 1 ) U ( 1 ,
)
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