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

question 49

Essay

For the functions $f ( x ) = \frac { 1 } { x - 2 }$ and $g ( x ) = \frac { x } { x - 1 }$ , find $f + g$ , $f - g$ , $\mathrm { fg }$ , $\frac { f } { g }$ and their domains. Answer $f + g = \frac { 1 } { x - 2 } + \frac { x } { x - 1 } = \frac { x ^ { 2 } - x - 1 } { ( x - 2 ) ( x - 1 ) } \quad$ domain: $( - \infty , 1 ) \cup ( 1,2 ) \cup ( 2 , \infty )$
$$\begin{array} { l l } f - g = \frac { 1 } { x - 2 } - \frac { x } { x - 1 } = - \frac { x ^ { 2 } - 3 x + 1 } { ( x - 2 ) ( x - 1 ) } & \text { domain: } ( - \infty , 1 ) \cup ( 1,2 ) \cup ( 2 , \infty ) \\f g = \left( \frac { 1 } { x - 2 } \right) \left( \frac { x } { x - 1 } \right) = \frac { x } { ( x - 1 ) ( x - 2 ) } & \text { domain: } ( - \infty , 1 ) \cup ( 1,2 ) \cup ( 2 , \infty ) \\\frac { f } { g } = \frac { ( x - 1 ) } { ( x - 2 ) x } & \text { domain: } ( - \infty , 0 ) \cup ( 0,1 ) \cup ( 1,2 ) \cup ( 2 , \infty )\end{array}$$ 10.
Given $f ( x ) = 2 x - 2$ and $g ( x ) = 2 x ^ { 2 }$ , find $( g \circ f ) ( - 1 )$ .

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