Let $f(x)=\frac x{x-2}$ . Find a function $y=g(x)$ so that $(f\circ g)(x)=4x$.
The solution is $g(x)=\frac{8x}{4x-1}$, but I don't know how to arrive at this answer. Please help me to find the steps to solve this problem.
Let $f(x)=\frac x{x-2}$ . Find a function $y=g(x)$ so that $(f\circ g)(x)=4x$.
The solution is $g(x)=\frac{8x}{4x-1}$, but I don't know how to arrive at this answer. Please help me to find the steps to solve this problem.
If $g(x)=y$ then $$4x=f(g(x))=f(y)=\frac{y}{y-2}$$ hence $4xy-8x=y$, from which you get $$y=\frac{8x}{4x-1}$$
Therefore $g(x)=y=\frac{8x}{4x-1}$.
Just write $(f\circ g)(x)$, which is $$\dfrac{g(x)}{g(x)-2}.$$ Now solve the equation $$\dfrac{g(x)}{g(x)-2}=4x$$ with respect to the variable $g(x)$.
$$f(x)=\frac { x }{ x-2 } ,y=g(x),(f\circ g)(x)=4x\\ \left( f\circ g \right) (x)=f\left( g\left( x \right) \right) =\frac { g\left( x \right) }{ g\left( x \right)-2 } =4x\Rightarrow \quad g\left( x \right)=4xg\left( x \right)-8x\Rightarrow g\left( x \right)\left( 4x-1 \right) =8x\\ \Rightarrow g\left( x \right)=\frac { 8x }{ 4x-1 } \\ $$
Ok. We know that $f(x)=\frac{x}{x-2}$
we also know that if we replace the variable x in f(x) with g(x) (let's call it "G" for simplicity) we must get 4x as a result. Let's write that down: $$\frac{G}{G-2}=4x$$ $$G=4x(G-2)$$ $$G=4xG-8x$$ $$G-4xG=-8x$$ $$G(1-4x)=-8x$$ $$G=\frac{-8x}{(1-4x)}$$ $$G=\frac{8x}{4x-1}$$