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For the Functions f(x)=1x2f ( x ) = \frac { 1 } { x - 2 }

question 49

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For the functions f(x)=1x2f ( x ) = \frac { 1 } { x - 2 } and g(x)=xx1g ( x ) = \frac { x } { x - 1 } , find f+gf + g , fgf - g , fg\mathrm { fg } , fg\frac { f } { g } and their domains. Answer f+g=1x2+xx1=x2x1(x2)(x1)f + g = \frac { 1 } { x - 2 } + \frac { x } { x - 1 } = \frac { x ^ { 2 } - x - 1 } { ( x - 2 ) ( x - 1 ) } \quad domain: (,1)(1,2)(2,)( - \infty , 1 ) \cup ( 1,2 ) \cup ( 2 , \infty )
fg=1x2xx1=x23x+1(x2)(x1) domain: (,1)(1,2)(2,)fg=(1x2)(xx1)=x(x1)(x2) domain: (,1)(1,2)(2,)fg=(x1)(x2)x domain: (,0)(0,1)(1,2)(2,)\begin{array} { l l } f - g = \frac { 1 } { x - 2 } - \frac { x } { x - 1 } = - \frac { x ^ { 2 } - 3 x + 1 } { ( x - 2 ) ( x - 1 ) } & \text { domain: } ( - \infty , 1 ) \cup ( 1,2 ) \cup ( 2 , \infty ) \\f g = \left( \frac { 1 } { x - 2 } \right) \left( \frac { x } { x - 1 } \right) = \frac { x } { ( x - 1 ) ( x - 2 ) } & \text { domain: } ( - \infty , 1 ) \cup ( 1,2 ) \cup ( 2 , \infty ) \\\frac { f } { g } = \frac { ( x - 1 ) } { ( x - 2 ) x } & \text { domain: } ( - \infty , 0 ) \cup ( 0,1 ) \cup ( 1,2 ) \cup ( 2 , \infty )\end{array} 10.
Given f(x)=2x2f ( x ) = 2 x - 2 and g(x)=2x2g ( x ) = 2 x ^ { 2 } , find (gf)(1)( g \circ f ) ( - 1 ) .

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