Question

Let `f(x) = 1 + 2x` and `g(x) = x/(x-1)`, how do you find each of the compositions and domain and range?

Answer 1

As explained below.

Explanation:

Composition f(g(x) = `1+2 (x/(x-1))` =`(3x-1)/(x-1)`

Its domain is x: `{R,x!= 1}`

For range, let f(g(x) =y= `(3x-1)/(x-1)`. now interchange x and y and the solve for y. Accordingly, x= `(3y-1)/(y-1)`. This gives xy-x=3y-1, so that y=`(x-1)/(x-3)`. The range of f(g(x) is x:`{R, x!= 3}`

Composition g(f(x) = `(1+2x)/(1+2x-1)` = `(1+2x)/(2x)`

Its domain is x: `{R, x !=0}`. For range, proceed as above by writing x= `(1+2y)/(2y) = 1 + 1/(2y)`, y=`1/(2(x-1)`

Hence Range of g(f(x) is `x: { R, x != 1}`