Q. 74 - Precalculus Enhanced with Graphing Utilities

Question

Given $f\left(x\right)=\frac{1}{x}$and $\left(\frac{f}{g}\right)\left(x\right)=\frac{x+1}{x^2-x}$, find the function $g$.

Step-by-Step Solution

Answer

The value of $\mathrm{g}\left(\mathrm{x}\right)=\frac{x-1}{x+1}$

Step 1. Given Information

Given $f\left(x\right)=\frac{1}{x}$ and $\left(\frac{f}{g}\right)\left(x\right)=\frac{x+1}{x^2-x}$,

We have to find the function $g$.

Step 2. We know that f g ( x ) = f ( x ) g ( x ) Simplifying the formula for g ( x )

Multiply with $g\left(x\right)$ on both side

$$\begin{gathered} \left(\frac{f}{g}\right)\left(x\right)·g\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}·g\left(x\right) \\ \left(\frac{f}{g}\right)\left(x\right)·g\left(x\right)=f\left(x\right) \\ \mathrm{Divide} \mathrm{by} \left(\frac{f}{g}\right)\left(x\right) \mathrm{on} \mathrm{both} \mathrm{side} \\ \left(\frac{\mathrm{f}}{\mathrm{g}}\right)\left(\mathrm{x}\right)·\mathrm{g}\left(\mathrm{x}\right)·\frac{1}{\left(\frac{\mathrm{f}}{\mathrm{g}}\right)\left(\mathrm{x}\right)}=\frac{\mathrm{f}\left(\mathrm{x}\right)}{\left(\frac{\mathrm{f}}{\mathrm{g}}\right)\left(\mathrm{x}\right)} \\ \mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{f}\left(\mathrm{x}\right)}{\left(\frac{\mathrm{f}}{\mathrm{g}}\right)\left(\mathrm{x}\right)} \end{gathered}$$

Step 3. Putting the value of f ( x )   and   f g ( x )

$$\begin{gathered} \mathrm{g}\left(\mathrm{x}\right)=\frac{\frac{1}{x}}{\frac{x+1}{x^2-x}} \\ \mathrm{g}\left(\mathrm{x}\right)=\frac{1}{x}·\frac{x^2-x}{x+1} \\ \mathrm{g}\left(\mathrm{x}\right)=\frac{1}{x}·\frac{x(x-1)}{x+1} \\ \mathrm{g}\left(\mathrm{x}\right)=\frac{x-1}{x+1} \end{gathered}$$