Function Composition

If $g(f(x))=\frac{x}{x+1}$ and $f(x)=\frac{1}{1-2x}$ , find $g(x)$ .

$\begin{equation*}g(f(x))=g(\frac{1}{1-2x})=\frac{x}{x+1}\end{equation}$

Sometimes you can do this type of question by inspection, but this one is a bit harder. I am going to use a variable substitution.

Let $u=\frac{1}{1-2x}$

$\begin{equation*}u=\frac{1}{1-2x}\end{equation}$

$\begin{equation*}1-2x=\frac{1}{u}\end{equation}$

$\begin{equation*}1-\frac{1}{u}=2x\end{equation}$

$\begin{equation*}\frac{u-1}{u}=2x\end{equation}$

$\begin{equation*}x=\frac{u-1}{2u}\end{equation}$

Therefore

$\begin{equation*}g(u)=\frac{\frac{u-1}{2u}}{\frac{u-1}{2u}+1}\end{equation}$

$\begin{equation*}g(u)=\frac{\frac{u-1}{2u}}{\frac{u-1+2u}{2u}}\end{equation}$

$\begin{equation*}g(u)=\frac{u-1}{3u-1}\end{equation}$

Therefore

$\begin{equation*}g(x)=\frac{x-1}{3x-1}\end{equation}$

Let’s test it

$g(x)=\frac{x-1}{3x-1}$ and $f(x)=\frac{1}{1-2x}$

$\begin{equation*}g(f(x))=\frac{\frac{1}{1-2x}-1}{\frac{3}{1-2x}-1}\end{equation}$

$\begin{equation*}g(f(x))=\frac{\frac{1-(1-2x)}{1-2x}}{\frac{3-(1-2x)}{1-2x}}\end{equation}$

$\begin{equation*}g(f(x))=\frac{2x}{2+2x}\end{equation}$

$\begin{equation*}g(f(x))=\frac{x}{x+1}\end{equation}$

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